Q.Integrate the following function: $\sqrt{x^2+4x+6}$

Showing the 12 most recent of 20 on this concept.

1. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
   Q.$\int \frac{dx}{\sqrt{(5+2x+x^2)^3}} =$ (A) $\frac{1}{4} \frac{1}{\sqrt{5+2x+x^2}} + C$ (B) $\frac{1}{\sqrt{5+2x+x^2}} + C$ (C) $\frac{x+1}{\sqrt{5+2x+x^2}} + C$ (D) $\frac{1}{4} \frac{x+1}{\sqrt{5+2x+x^2}} + C$

   ›Reveal solutionSolution

   We simplify the quadratic expression by completing the square, then use a trigonometric substitution to evaluate the integral, finding the result $\frac{1}{4} \frac{x+1}{\sqrt{5+2x+x^2}} + C$.

   The integral involves a quadratic expression $5+2x+x^2$ raised to the power of $3/2$ in the denominator. Integrals of this form, especially those with quadratic expressions under a square root, are typically solved by first completing the square in the quadratic expression. This transforms the quadratic into a sum or difference of squares, which then suggests a standard trigonometric or hyperbolic substitution to simplify the integral.

   Here's how to solve it step-by-step:

   1. Complete the square for the quadratic expression:

      The quadratic expression in the denominator is $x^2+2x+5$. To complete the square, we look for a term $(x+k)^2 = x^2+2kx+k^2$.

      Comparing $x^2+2x$ with $x^2+2kx$, we see $2k=2$, so $k=1$.

      Thus, $(x+1)^2 = x^2+2x+1$.

      We can rewrite $x^2+2x+5$ as $(x^2+2x+1) + 4$.

      So, $x^2+2x+5 = (x+1)^2 + 2^2$.

      Completing the square: $ax^2+bx+c = a\left(x+\frac{b}{2a}\right)^2 + \left(c-\frac{b^2}{4a}\right)$

      The integral now becomes:

   $\int \frac{dx}{\sqrt{((x+1)^2 + 2^2)^3}} = \int \frac{dx}{((x+1)^2 + 2^2)^{3/2}}$

   2. Choose the appropriate trigonometric substitution:

      The expression is in the form $(u^2+a^2)$, where $u = x+1$ and $a=2$. For expressions of the form $\sqrt{u^2+a^2}$ or $(u^2+a^2)^{n/2}$, the standard trigonometric substitution is $u = a \tan \theta$.

      Let $x+1 = 2 \tan \theta$.

   3. Calculate $dx$ and substitute into the integral:

      Differentiate both sides of $x+1 = 2 \tan \theta$ with respect to $\theta$:

      $dx = 2 \sec^2 \theta \, d\theta$.

      Now, substitute $x+1 = 2 \tan \theta$ and $dx = 2 \sec^2 \theta \, d\theta$ into the integral:

      The denominator becomes:

   $((2 \tan \theta)^2 + 2^2)^{3/2} = (4 \tan^2 \theta + 4)^{3/2}$

   $= (4(\tan^2 \theta + 1))^{3/2}$

   Using the identity $\tan^2 \theta + 1 = \sec^2 \theta$:
   

   $= (4 \sec^2 \theta)^{3/2}$

   $= (2^2 \sec^2 \theta)^{3/2} = (2 \sec \theta)^3 = 8 \sec^3 \theta$

   > [!WARNING]
   > Be careful when simplifying $(A^2)^{3/2}$. It is $(A^2)^{3/2} = A^3$. Here, $A = 2 \sec \theta$, so $(4 \sec^2 \theta)^{3/2} = (2 \sec \theta)^3 = 8 \sec^3 \theta$. A common mistake is to miscalculate the power.
   
   Substitute these back into the integral:
   

   $\int \frac{2 \sec^2 \theta \, d\theta}{8 \sec^3 \theta}$

   Simplify the expression:
   

   $\int \frac{1}{4 \sec \theta} \, d\theta = \frac{1}{4} \int \frac{1}{\sec \theta} \, d\theta$

   Since $\frac{1}{\sec \theta} = \cos \theta$:
   

   $\frac{1}{4} \int \cos \theta \, d\theta$

   4. Evaluate the trigonometric integral: The integral of $\cos \theta$ is $\sin \theta$.

   $\frac{1}{4} \sin \theta + C$

   5. Convert the result back to the original variable $x$:

      We need to express $\sin \theta$ in terms of $x$. From our substitution, we have $x+1 = 2 \tan \theta$, which means $\tan \theta = \frac{x+1}{2}$.

      We can construct a right-angled triangle where $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$.

      Let the opposite side be $x+1$ and the adjacent side be $2$.

      Using the Pythagorean theorem, the hypotenuse is $\sqrt{(\text{opposite})^2 + (\text{adjacent})^2}$:

      Hypotenuse $= \sqrt{(x+1)^2 + 2^2} = \sqrt{x^2+2x+1+4} = \sqrt{x^2+2x+5}$.

      Now, we can find $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$:

   $\sin \theta = \frac{x+1}{\sqrt{x^2+2x+5}}$

   Substitute this back into our integrated expression:
   

   $\frac{1}{4} \left( \frac{x+1}{\sqrt{x^2+2x+5}} \right) + C$

   $= \frac{1}{4} \frac{x+1}{\sqrt{5+2x+x^2}} + C$

   Comparing this result with the given options, it matches option (D).

   ✓Final answer

   The value of the integral is $\boxed{\frac{1}{4} \frac{x+1}{\sqrt{5+2x+x^2}} + C}$.

2. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.$\int \frac{dx}{(x^2+1)(x^2+4)} =$ (A) $\frac{1}{3} \tan^{-1} x + \frac{1}{6} \tan^{-1} \left(\frac{x}{2}\right) + c$ (B) $\frac{1}{3} \tan^{-1} x - \frac{1}{3} \tan^{-1} \left(\frac{x}{2}\right) + c$ (C) $\frac{1}{3} \tan^{-1} x + \frac{1}{3} \tan^{-1} \left(\frac{x}{2}\right) + c$ (D) $\frac{1}{3} \tan^{-1} x - \frac{1}{6} \tan^{-1} \left(\frac{x}{2}\right) + c$

   ›Reveal solutionSolution

   The integral is solved by partial fractions after factoring the denominator into a sum of simpler rational terms, then integrating each to an arctangent. The result is $\frac{1}{3}\tan^{-1}x - \frac{1}{6}\tan^{-1}\left(\frac{x}{2}\right) + c$, which corresponds to option (D).

   The key idea is that the denominator is a product of two irreducible quadratics: $x^2+1$ and $x^2+4$. There is no linear factor, so the standard partial fraction decomposition for such a case uses numerators of the form $Ax+B$ and $Cx+D$. But here, because both quadratics are of the form $x^2 + a^2$, we can use a clever shortcut: the integrand can be split into a difference of two simpler fractions by noting that $(x^2+4) - (x^2+1) = 3$. This lets us avoid solving a system of equations.

   1. Set up the partial fraction decomposition We want constants $A, B, C, D$ such that

   $\frac{1}{(x^2+1)(x^2+4)} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+4}.$

   Multiply through by the denominator:

   $1 = (Ax+B)(x^2+4) + (Cx+D)(x^2+1).$

   2. Expand and collect like terms

   $1 = (A+C)x^3 + (B+D)x^2 + (4A + C)x + (4B + D).$

   For this to hold for all $x$, coefficients of $x^3$, $x^2$, $x$, and the constant must match.

   • $x^3$: $A + C = 0$
   • $x^2$: $B + D = 0$
   • $x$: $4A + C = 0$
   • constant: $4B + D = 1$

   3. Solve the system From $A + C = 0$ we have $C = -A$. Substitute into $4A + C = 0$:

   $4A - A = 0 \quad\Rightarrow\quad 3A = 0 \quad\Rightarrow\quad A = 0,\; C = 0.$

   From $B + D = 0$ we have $D = -B$. Substitute into $4B + D = 1$:

   $4B - B = 1 \quad\Rightarrow\quad 3B = 1 \quad\Rightarrow\quad B = \frac{1}{3},\; D = -\frac{1}{3}.$

   So the decomposition is:

   $\frac{1}{(x^2+1)(x^2+4)} = \frac{1/3}{x^2+1} - \frac{1/3}{x^2+4}.$

   4. Integrate term by term Recall that $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + c$.

   $\int \frac{1/3}{x^2+1}\,dx = \frac{1}{3} \tan^{-1} x + c_1,$

   $\int \frac{1/3}{x^2+4}\,dx = \frac{1}{3} \cdot \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + c_2 = \frac{1}{6} \tan^{-1}\left(\frac{x}{2}\right) + c_2.$

   Therefore,

   $\int \frac{dx}{(x^2+1)(x^2+4)} = \frac{1}{3} \tan^{-1} x - \frac{1}{6} \tan^{-1}\left(\frac{x}{2}\right) + c.$

   Tip

   A faster method: notice that $\frac{1}{(x^2+1)(x^2+4)} = \frac{1}{3}\left(\frac{1}{x^2+1} - \frac{1}{x^2+4}\right)$ because $(x^2+4) - (x^2+1) = 3$. This gives the decomposition instantly without solving any system.

   Watch out

   A common mistake is to forget the factor $\frac{1}{a}$ when integrating $\frac{1}{x^2+a^2}$. For $a=2$, the antiderivative is $\frac{1}{2}\tan^{-1}(x/2)$, not $\tan^{-1}(x/2)$.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

3. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.$\int \frac{\cos 2x + \sin 4x}{\sqrt{3 \sin 2x - 2}} dx =$ (A) $\frac{1}{27} \sqrt{3 \sin 2x - 2} (6 \sin 2x + 17) + c$ (B) $\frac{\sqrt{3 \sin 2x - 2}}{27 (6 \sin 2x + 17)} + c$ (C) $\frac{27 (6 \sin 2x + 17)}{\sqrt{3 \sin 2x - 2}} + c$ (D) $27 \sqrt{3 \sin 2x - 2} (6 \sin 2x + 17) + c$

   ›Reveal solutionSolution

   The integral simplifies by rewriting the numerator in terms of $\sin 2x$ and using a substitution $t = \sin 2x$, leading to a rational function in $t$ that integrates to $\frac{1}{27} \sqrt{3 \sin 2x - 2} (6 \sin 2x + 17) + c$, which matches option (A).

   The key insight here is that the denominator $\sqrt{3 \sin 2x - 2}$ suggests a substitution $u = 3 \sin 2x - 2$, but the numerator contains $\cos 2x$ and $\sin 4x$. Notice that $\sin 4x = 2 \sin 2x \cos 2x$, so the whole numerator factors as $\cos 2x (1 + 2 \sin 2x)$. That $\cos 2x$ is exactly the derivative of $\sin 2x$ up to a constant factor, which makes a substitution in terms of $\sin 2x$ natural. Once we express everything in $t = \sin 2x$, the integral becomes a straightforward rational function integration.

   1. Rewrite the numerator Use $\sin 4x = 2 \sin 2x \cos 2x$:

   $\cos 2x + \sin 4x = \cos 2x + 2 \sin 2x \cos 2x = \cos 2x (1 + 2 \sin 2x).$

   The integral becomes

   $\int \frac{\cos 2x (1 + 2 \sin 2x)}{\sqrt{3 \sin 2x - 2}} \, dx.$

   2. Substitute $t = \sin 2x$ Then $dt = 2 \cos 2x \, dx$, so $\cos 2x \, dx = \frac{dt}{2}$. The integral transforms to

   $\int \frac{(1 + 2t)}{\sqrt{3t - 2}} \cdot \frac{dt}{2} = \frac12 \int \frac{1 + 2t}{\sqrt{3t - 2}} \, dt.$

   3. Simplify the integrand Let $u = 3t - 2$, so $t = \frac{u+2}{3}$ and $dt = \frac{du}{3}$. Then $1 + 2t = 1 + 2 \cdot \frac{u+2}{3} = 1 + \frac{2u+4}{3} = \frac{3 + 2u + 4}{3} = \frac{2u + 7}{3}$. The integral becomes

   $\frac12 \int \frac{\frac{2u+7}{3}}{\sqrt{u}} \cdot \frac{du}{3} = \frac12 \cdot \frac13 \cdot \frac13 \int \frac{2u+7}{\sqrt{u}} \, du = \frac{1}{18} \int (2u^{1/2} + 7u^{-1/2}) \, du.$

   4. Integrate term by term

   $\frac{1}{18} \left( 2 \cdot \frac{u^{3/2}}{3/2} + 7 \cdot \frac{u^{1/2}}{1/2} \right) = \frac{1}{18} \left( \frac{4}{3} u^{3/2} + 14 u^{1/2} \right) = \frac{1}{18} \cdot \frac{4u^{3/2} + 42 u^{1/2}}{3} = \frac{4u^{3/2} + 42 u^{1/2}}{54}.$

   Simplify by factoring $2u^{1/2}$:

   $\frac{2u^{1/2} (2u + 21)}{54} = \frac{u^{1/2} (2u + 21)}{27}.$

   5. Back-substitute Recall $u = 3t - 2 = 3 \sin 2x - 2$, so $u^{1/2} = \sqrt{3 \sin 2x - 2}$ and $2u + 21 = 2(3 \sin 2x - 2) + 21 = 6 \sin 2x - 4 + 21 = 6 \sin 2x + 17$. Hence the integral is

   $\frac{1}{27} \sqrt{3 \sin 2x - 2} \, (6 \sin 2x + 17) + c.$

   Watch out

   A common mistake is forgetting the factor $\frac12$ from $dt = 2 \cos 2x \, dx$, or mishandling the substitution $u = 3t - 2$ — double-check each algebraic step to avoid sign errors.

   ✓Final answer

   The correct option is (A): $\frac{1}{27} \sqrt{3 \sin 2x - 2} (6 \sin 2x + 17) + c$.

4. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.For $x > 0$ $\int \left( \dfrac{\sqrt{1+x+x^2}}{1+x} + \dfrac{1}{2\sqrt{1+x+x^2}} - \dfrac{1}{(1+x)\sqrt{1+x+x^2}} \right) dx =$ (A) $\dfrac{1}{\sqrt{1+x+x^2}} + C$ (B) $\sqrt{1+x} + C$ (C) $\dfrac{1}{\sqrt{1+x}} + C$ (D) $\sqrt{x^2 + x + 1} + C$

   ›Reveal solutionSolution

   The integrand is the derivative of $\sqrt{1+x+x^2}$ with respect to $x$, so the integral is $\sqrt{x^2+x+1}+C$, which is option (D).

   The problem gives a sum of three terms inside an integral. When you see a complicated-looking integrand like this, the first instinct should not be to integrate each term separately by brute force. Instead, look for a pattern: the integrand might be the result of a derivative you already know.

   Notice that $\sqrt{1+x+x^2}$ is a function whose derivative involves $\frac{1}{2\sqrt{1+x+x^2}}$ times the derivative of the inside, $1+2x$. That derivative is

   $\frac{d}{dx}\sqrt{1+x+x^2} = \frac{1+2x}{2\sqrt{1+x+x^2}}.$

   Our integrand has a term $\frac{1}{2\sqrt{1+x+x^2}}$ but also other pieces. The key is to see if the whole sum can be written as that derivative, perhaps after splitting the numerator $1+2x$ into parts that match the other terms.

   Let’s check step by step.

   1. Write the derivative we suspect:

   $\frac{d}{dx}\sqrt{1+x+x^2} = \frac{1+2x}{2\sqrt{1+x+x^2}}.$

   We want to see if the given integrand equals this.

   2. The given integrand is

   $\frac{\sqrt{1+x+x^2}}{1+x} + \frac{1}{2\sqrt{1+x+x^2}} - \frac{1}{(1+x)\sqrt{1+x+x^2}}.$

   Combine the first and third terms, since they share the factor $\frac{1}{1+x}$:

   $\frac{\sqrt{1+x+x^2}}{1+x} - \frac{1}{(1+x)\sqrt{1+x+x^2}} = \frac{1}{1+x}\left(\sqrt{1+x+x^2} - \frac{1}{\sqrt{1+x+x^2}}\right).$

   3. Put everything over a common denominator $\sqrt{1+x+x^2}$ inside the parentheses:

   $\sqrt{1+x+x^2} - \frac{1}{\sqrt{1+x+x^2}} = \frac{(1+x+x^2) - 1}{\sqrt{1+x+x^2}} = \frac{x+x^2}{\sqrt{1+x+x^2}}.$

   So the first and third terms together become

   $\frac{1}{1+x} \cdot \frac{x+x^2}{\sqrt{1+x+x^2}} = \frac{x(1+x)}{(1+x)\sqrt{1+x+x^2}} = \frac{x}{\sqrt{1+x+x^2}}.$

   4. Now add the remaining middle term $\frac{1}{2\sqrt{1+x+x^2}}$:

   $\frac{x}{\sqrt{1+x+x^2}} + \frac{1}{2\sqrt{1+x+x^2}} = \frac{2x + 1}{2\sqrt{1+x+x^2}}.$

   5. That is exactly $\frac{1+2x}{2\sqrt{1+x+x^2}}$, which is the derivative of $\sqrt{1+x+x^2}$.

   Therefore the integrand is the derivative of $\sqrt{x^2+x+1}$.

   Watch out

   A common mistake is to try integrating each term separately using substitution or trigonometric methods. That is unnecessarily long and error-prone. Recognizing the derivative pattern saves time and avoids algebra mistakes.

   Since the integral of a derivative is the original function plus a constant,

   $\int \left( \frac{\sqrt{1+x+x^2}}{1+x} + \frac{1}{2\sqrt{1+x+x^2}} - \frac{1}{(1+x)\sqrt{1+x+x^2}} \right) dx = \sqrt{x^2+x+1} + C.$

   ✓Final answer

   The correct option is (D): $\sqrt{x^2+x+1}+C$.

5. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.

   [!FORMULA] $\int \frac{\sin^2 x \tan x}{\cos^6 x + \sin^4 x \cos^2 x + \cos^4 x \sin^2 x + \sin^6 x} dx =$

   (A) $\log(\sin^4 x + \cos^4 x) + c$ (B) $\frac{1}{4} \log(\sin^4 x + \cos^4 x) + c$ (C) $\frac{1}{4} \log(1 + \tan^4 x) + c$ (D) $\log(1 + \tan^4 x) + c$

   ›Reveal solutionSolution

   The integrand simplifies dramatically by factoring the denominator as a sum of two squares, leading to a clean substitution that yields $\frac{1}{4}\log(\sin^4 x + \cos^4 x) + c$, which matches option (B).

   ---

   The key insight is that the denominator looks messy but is actually a disguised sum of two perfect squares. Once we see that, the numerator also cooperates, and a simple substitution finishes the job.

   Why this works:

   The denominator has terms $\cos^6 x$, $\sin^6 x$, and mixed terms $\sin^4 x \cos^2 x$, $\cos^4 x \sin^2 x$. This suggests grouping as $(\cos^6 x + \sin^6 x) + \sin^2 x \cos^2 x (\sin^2 x + \cos^2 x)$. Since $\sin^2 x + \cos^2 x = 1$, the denominator becomes $\cos^6 x + \sin^6 x + \sin^2 x \cos^2 x$. And $\cos^6 x + \sin^6 x$ itself factors as $(\cos^2 x + \sin^2 x)(\cos^4 x - \sin^2 x \cos^2 x + \sin^4 x) = \cos^4 x - \sin^2 x \cos^2 x + \sin^4 x$. Adding the extra $\sin^2 x \cos^2 x$ gives exactly $\cos^4 x + \sin^4 x$. That’s the clean core.

   Now the numerator $\sin^2 x \tan x = \sin^2 x \cdot \frac{\sin x}{\cos x} = \frac{\sin^3 x}{\cos x}$. So the whole integrand becomes $\frac{\sin^3 x}{\cos x (\sin^4 x + \cos^4 x)}$. A substitution $t = \sin^4 x + \cos^4 x$ will work because its derivative involves $\sin^3 x \cos x$ — almost what we have, except we have $\frac{\sin^3 x}{\cos x}$. A small adjustment with $\cos^2 x$ fixes it.

   Let’s go step by step.

   1. Simplify the denominator

   $D = \cos^6 x + \sin^4 x \cos^2 x + \cos^4 x \sin^2 x + \sin^6 x$

   Group as $(\cos^6 x + \sin^6 x) + \sin^2 x \cos^2 x (\sin^2 x + \cos^2 x)$.

   Since $\sin^2 x + \cos^2 x = 1$, we have

   $D = \cos^6 x + \sin^6 x + \sin^2 x \cos^2 x.$

   Now use the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ with $a = \cos^2 x$, $b = \sin^2 x$:

   $\cos^6 x + \sin^6 x = (\cos^2 x + \sin^2 x)(\cos^4 x - \sin^2 x \cos^2 x + \sin^4 x) = \cos^4 x - \sin^2 x \cos^2 x + \sin^4 x.$

   Adding the leftover $\sin^2 x \cos^2 x$ cancels the middle term:

   $D = \cos^4 x + \sin^4 x.$

   So the denominator is simply $\sin^4 x + \cos^4 x$.

   2. Rewrite the integrand The numerator is $\sin^2 x \tan x = \sin^2 x \cdot \frac{\sin x}{\cos x} = \frac{\sin^3 x}{\cos x}$. Hence the integral becomes

   $I = \int \frac{\sin^3 x}{\cos x \, (\sin^4 x + \cos^4 x)} \, dx.$

   3. Choose a substitution Let $u = \sin^4 x + \cos^4 x$. Then

   $du = (4 \sin^3 x \cos x - 4 \cos^3 x \sin x) \, dx = 4 \sin x \cos x (\sin^2 x - \cos^2 x) \, dx.$

   That’s not directly our numerator. Instead, try $t = \sin^4 x + \cos^4 x$ but multiply numerator and denominator by $\cos x$ to get $\sin^3 x \cos x$ in the numerator.

   Actually, a better approach: multiply numerator and denominator by $\cos x$:

   $I = \int \frac{\sin^3 x \cos x}{\cos^2 x (\sin^4 x + \cos^4 x)} \, dx = \int \frac{\sin^3 x \cos x}{(1 - \sin^2 x)(\sin^4 x + \cos^4 x)} \, dx.$

   That’s messy. Instead, note that $\sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x$, but that doesn’t help directly.

   A cleaner substitution: let $t = \sin^4 x + \cos^4 x$. Compute $dt$ differently:

   $\frac{d}{dx}(\sin^4 x + \cos^4 x) = 4\sin^3 x \cos x - 4\cos^3 x \sin x = 4\sin x \cos x (\sin^2 x - \cos^2 x).$

   Not matching. But we can also write

   $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2 2x.$

   Still not helpful.

   The real trick: rewrite the integrand as

   $\frac{\sin^3 x}{\cos x (\sin^4 x + \cos^4 x)} = \frac{\sin^3 x \cos x}{\cos^2 x (\sin^4 x + \cos^4 x)} = \frac{\sin^3 x \cos x}{(1 - \sin^2 x)(\sin^4 x + \cos^4 x)}.$

   That’s not simpler. Instead, try $u = \sin^2 x$? Then $du = 2\sin x \cos x \, dx$, and $\sin^3 x dx = \sin x \cdot \sin^2 x dx = \sin x \cdot u \, dx$, but we have $\frac{\sin^3 x}{\cos x} dx = \frac{\sin x \cdot \sin^2 x}{\cos x} dx = \frac{\sin x \cdot u}{\cos x} dx$. Not a clean match.

   Let’s step back. Multiply numerator and denominator by $\cos x$:

   $I = \int \frac{\sin^3 x \cos x}{\cos^2 x (\sin^4 x + \cos^4 x)} dx.$

   Now note $\cos^2 x = 1 - \sin^2 x$, but better: write $\sin^3 x \cos x = \frac{1}{4} \cdot 4 \sin^3 x \cos x$. Observe that

   $\frac{d}{dx}(\sin^4 x) = 4 \sin^3 x \cos x.$

   So $\sin^3 x \cos x \, dx = \frac{1}{4} d(\sin^4 x)$. Also $\cos^2 x = 1 - \sin^2 x$, but we still have $\sin^4 x + \cos^4 x$ in denominator. Write $\cos^4 x = (1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x$. Then

   $\sin^4 x + \cos^4 x = \sin^4 x + 1 - 2\sin^2 x + \sin^4 x = 2\sin^4 x - 2\sin^2 x + 1.$

   That’s quadratic in $\sin^2 x$. Let $u = \sin^2 x$, then $du = 2\sin x \cos x \, dx$, and $\sin^3 x \cos x \, dx = \sin^2 x \cdot \sin x \cos x \, dx = u \cdot \frac{du}{2}$. Also $\cos^2 x = 1 - u$. So

   $I = \int \frac{u \cdot \frac{du}{2}}{(1-u)(2u^2 - 2u + 1)} = \frac{1}{2} \int \frac{u}{(1-u)(2u^2 - 2u + 1)} du.$

   This is doable but messy. There must be a simpler way.

   4. The elegant substitution Notice that $\frac{d}{dx}(\sin^4 x + \cos^4 x) = 4\sin^3 x \cos x - 4\cos^3 x \sin x = 4\sin x \cos x (\sin^2 x - \cos^2 x)$. Our numerator is $\frac{\sin^3 x}{\cos x} dx$. Multiply numerator and denominator by $\cos x$ to get $\frac{\sin^3 x \cos x}{\cos^2 x} dx$. But we want $4\sin^3 x \cos x$ for the derivative. So write

   $I = \int \frac{\sin^3 x}{\cos x (\sin^4 x + \cos^4 x)} dx = \int \frac{4\sin^3 x \cos x}{4\cos^2 x (\sin^4 x + \cos^4 x)} dx.$

   Now $4\sin^3 x \cos x \, dx = d(\sin^4 x)$. But we have $\cos^2 x$ in denominator. Write $\cos^2 x = 1 - \sin^2 x$. Not great.

   Alternatively, use the identity $\sin^4 x + \cos^4 x = \frac{1}{2}(1 + \cos^2 2x)$? Actually $\sin^4 x + \cos^4 x = 1 - \frac{1}{2}\sin^2 2x = \frac{3}{4} + \frac{1}{4}\cos 4x$. That might lead to a tangent substitution.

   Let’s try dividing numerator and denominator by $\cos^4 x$:

   $\frac{\sin^2 x \tan x}{\sin^4 x + \cos^4 x} = \frac{\sin^2 x \cdot \frac{\sin x}{\cos x}}{\cos^4 x (\tan^4 x + 1)} = \frac{\sin^3 x}{\cos^5 x} \cdot \frac{1}{1 + \tan^4 x}.$

   But $\frac{\sin^3 x}{\cos^5 x} = \tan^3 x \cdot \sec^2 x$. And $\sec^2 x \, dx = d(\tan x)$. So

   $I = \int \frac{\tan^3 x}{1 + \tan^4 x} \cdot \sec^2 x \, dx = \int \frac{\tan^3 x}{1 + \tan^4 x} \, d(\tan x).$

   Let $t = \tan x$, then $dt = \sec^2 x \, dx$, and

   $I = \int \frac{t^3}{1 + t^4} \, dt.$

   This is a standard integral: let $u = 1 + t^4$, then $du = 4t^3 dt$, so $t^3 dt = \frac{du}{4}$. Hence

   $I = \int \frac{1}{u} \cdot \frac{du}{4} = \frac{1}{4} \log |u| + c = \frac{1}{4} \log(1 + t^4) + c = \frac{1}{4} \log(1 + \tan^4 x) + c.$

   That’s option (C). But wait — check the original denominator: we had $\sin^4 x + \cos^4 x$ in denominator after simplification, and dividing by $\cos^4 x$ gives $\tan^4 x + 1$, yes. So the integral is $\frac{1}{4} \log(1 + \tan^4 x) + c$. That matches (C).

   However, note that $\sin^4 x + \cos^4 x = \cos^4 x (1 + \tan^4 x)$, so $\log(1 + \tan^4 x) = \log(\sin^4 x + \cos^4 x) - 4\log|\cos x|$. The constant from $\log|\cos x|$ can be absorbed into $c$ only if it’s not there — but here it’s not a constant. So (C) and (B) are not the same unless we check carefully. Let’s verify:

   $\frac{1}{4}\log(1+\tan^4 x) = \frac{1}{4}\log\left(\frac{\sin^4 x + \cos^4 x}{\cos^4 x}\right) = \frac{1}{4}\log(\sin^4 x + \cos^4 x) - \log|\cos x|.$

   That extra $-\log|\cos x|$ means (C) and (B) differ by a non-constant term. So which is correct? Let’s differentiate both candidates.

   Differentiate $\frac{1}{4}\log(\sin^4 x + \cos^4 x)$:

   $\frac{1}{4} \cdot \frac{4\sin^3 x \cos x - 4\cos^3 x \sin x}{\sin^4 x + \cos^4 x} = \frac{\sin x \cos x (\sin^2 x - \cos^2 x)}{\sin^4 x + \cos^4 x}.$

   That’s not our integrand $\frac{\sin^3 x}{\cos x (\sin^4 x + \cos^4 x)}$. So (B) is not correct.

   Differentiate $\frac{1}{4}\log(1+\tan^4 x)$:

   $\frac{1}{4} \cdot \frac{4\tan^3 x \sec^2 x}{1+\tan^4 x} = \frac{\tan^3 x \sec^2 x}{1+\tan^4 x}.$

   But $\tan^3 x \sec^2 x = \frac{\sin^3 x}{\cos^3 x} \cdot \frac{1}{\cos^2 x} = \frac{\sin^3 x}{\cos^5 x}$. And $1+\tan^4 x = \frac{\sin^4 x + \cos^4 x}{\cos^4 x}$. So the derivative becomes

   $\frac{\sin^3 x}{\cos^5 x} \cdot \frac{\cos^4 x}{\sin^4 x + \cos^4 x} = \frac{\sin^3 x}{\cos x (\sin^4 x + \cos^4 x)},$

   which matches exactly. So (C) is correct.

   Watch out

   A common mistake is to stop at $\frac{1}{4}\log(\sin^4 x + \cos^4 x)$ because it looks neat, but the derivative doesn’t match — the missing $\cos x$ in the numerator changes everything. Always differentiate to check.

   Tip

   Dividing numerator and denominator by $\cos^4 x$ converts the denominator into $1+\tan^4 x$ and the numerator into $\tan^3 x \sec^2 x$, which is a perfect differential for $t = \tan x$. This is the cleanest path.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

6. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.$\int (x+2)\sqrt{x+3}\ dx =$ (A) $\dfrac{2}{15}\sqrt{x+3}\left(3x^2-13x+12\right)+C$ (B) $\dfrac{2}{15}\sqrt{x+3}\left(3x^2+13x+12\right)+C$ (C) $\dfrac{2}{5}\sqrt{x+3}\left(3x^2-12x+13\right)+C$ (D) $\dfrac{2}{5}\sqrt{x+3}\left(3x^2+12x+13\right)+C$

   ›Reveal solutionSolution

   The integral is solved by substituting $t = \sqrt{x+3}$, which turns the integrand into a polynomial in $t$. After integrating and back-substituting, the result matches option (B).

   The key insight: when you see a linear expression inside a square root, the substitution $t = \sqrt{\text{that linear expression}}$ often works beautifully. Here, $\sqrt{x+3}$ is the stubborn part — so let it become the new variable. This transforms the integral into a simple polynomial integration, avoiding messy expansion or integration by parts.

   1. Set up the substitution.

      Let $t = \sqrt{x+3}$. Then $t^2 = x+3$, so $x = t^2 - 3$.

      Differentiating: $dx = 2t\, dt$.

   2. Rewrite the integrand in terms of $t$.

      The factor $(x+2)$ becomes $(t^2 - 3 + 2) = t^2 - 1$.

      The factor $\sqrt{x+3}$ is simply $t$.

      So the integral becomes:

   $\int (t^2 - 1) \cdot t \cdot (2t\, dt) = \int (t^2 - 1)(2t^2)\, dt = \int (2t^4 - 2t^2)\, dt.$

   3. Integrate term by term.

   $\int 2t^4\, dt = \frac{2t^5}{5}, \quad \int -2t^2\, dt = -\frac{2t^3}{3}.$

   So the indefinite integral is:

   $\frac{2t^5}{5} - \frac{2t^3}{3} + C.$

   4. Factor out a common factor to match the answer format. Notice the options have a single factor $\sqrt{x+3}$ times a quadratic in $x$. So factor $\frac{2t^3}{15}$ (since $t = \sqrt{x+3}$):

   $\frac{2t^3}{15} \left(3t^2 - 5\right) + C.$

   Check: $\frac{2t^3}{15} \cdot 3t^2 = \frac{2t^5}{5}$, and $\frac{2t^3}{15} \cdot (-5) = -\frac{2t^3}{3}$. Yes.

   5. Back-substitute $t = \sqrt{x+3}$. $t^3 = (x+3)^{3/2} = (x+3)\sqrt{x+3}$, and $t^2 = x+3$. So:

   $\frac{2}{15} (x+3)\sqrt{x+3} \left[3(x+3) - 5\right] + C.$

   Simplify the bracket: $3(x+3) - 5 = 3x + 9 - 5 = 3x + 4$.

   6. Expand to get the quadratic in $x$.

   $\frac{2}{15} \sqrt{x+3} \left[(x+3)(3x+4)\right] + C.$

   Multiply: $(x+3)(3x+4) = 3x^2 + 4x + 9x + 12 = 3x^2 + 13x + 12$.

   Therefore:

   $\int (x+2)\sqrt{x+3}\, dx = \frac{2}{15} \sqrt{x+3} \left(3x^2 + 13x + 12\right) + C.$

   Watch out

   A common mistake is to forget the factor $2t$ from $dx = 2t\, dt$, or to mishandle the algebra when expanding $(x+3)(3x+4)$. Double-check the expansion: $3x^2 + 13x + 12$ is correct.

   ✓Final answer

   The correct option is (B).

7. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.$\int \frac{2x+3}{\sqrt{3x^2-2x+1}} \, dx =$ (A) $\frac{2}{3} \sqrt{3x^2-2x+1} + \frac{11}{3} \sin^{-1} \left( \frac{3x-1}{\sqrt{2}} \right) + c$ (B) $\frac{1}{3} \sqrt{3x^2-2x+1} + \frac{11}{3} \sin^{-1} \left( \frac{\sqrt{3}x-1}{\sqrt{2}} \right) + c$ (C) $\frac{1}{3} \sqrt{3x^2-2x+1} + \frac{11}{3} \sin^{-1} \left( \frac{3x-1}{\sqrt{3}} \right) + c$ (D) $\frac{2}{3} \sqrt{3x^2-2x+1} + \frac{11}{3\sqrt{3}} \sin^{-1} \left( \frac{3x-1}{\sqrt{2}} \right) + c$

   ›Reveal solutionSolution

   The integral splits into a derivative‑matching part (giving a square root term) and a constant part (giving an inverse sine). After completing the square and integrating, the result matches option (D).

   ---

   We want

   $\int \frac{2x+3}{\sqrt{3x^2-2x+1}}\,dx.$

   The denominator’s radicand is a quadratic; its derivative is $6x-2$. The numerator $2x+3$ is almost a multiple of that derivative. This suggests we write the numerator as

   $2x+3 = A(6x-2) + B,$

   so that the integral splits into a part where the numerator is exactly the derivative of the radicand (giving a simple substitution) and a constant part that leads to an inverse trigonometric form after completing the square.

   ---

   1. Find constants $A$ and $B$.

   $2x+3 = A(6x-2) + B = 6A\,x + (-2A + B).$

   Equate coefficients:

   $6A = 2 \quad\Rightarrow\quad A = \frac{1}{3},$

   $-2A + B = 3 \quad\Rightarrow\quad -\frac{2}{3} + B = 3 \quad\Rightarrow\quad B = \frac{11}{3}.$

   So

   $2x+3 = \frac{1}{3}(6x-2) + \frac{11}{3}.$

   2. Split the integral.

   $\int \frac{2x+3}{\sqrt{3x^2-2x+1}}\,dx = \frac{1}{3}\int \frac{6x-2}{\sqrt{3x^2-2x+1}}\,dx + \frac{11}{3}\int \frac{dx}{\sqrt{3x^2-2x+1}}.$

   3. First integral: direct substitution. Let $u = 3x^2-2x+1$, so $du = (6x-2)\,dx$. Then

   $\frac{1}{3}\int \frac{du}{\sqrt{u}} = \frac{1}{3}\cdot 2\sqrt{u} = \frac{2}{3}\sqrt{3x^2-2x+1}.$

   4. Second integral: complete the square.

   $3x^2-2x+1 = 3\left(x^2 - \frac{2}{3}x\right) + 1 = 3\left[\left(x - \frac{1}{3}\right)^2 - \frac{1}{9}\right] + 1 = 3\left(x - \frac{1}{3}\right)^2 - \frac{1}{3} + 1 = 3\left(x - \frac{1}{3}\right)^2 + \frac{2}{3}.$

   Factor out the 3:

   $= 3\left[\left(x - \frac{1}{3}\right)^2 + \frac{2}{9}\right].$

   So

   $\sqrt{3x^2-2x+1} = \sqrt{3}\,\sqrt{\left(x - \frac{1}{3}\right)^2 + \left(\frac{\sqrt{2}}{3}\right)^2}.$

   5. Integrate the constant part.

   $\frac{11}{3}\int \frac{dx}{\sqrt{3}\,\sqrt{(x - 1/3)^2 + (\sqrt{2}/3)^2}} = \frac{11}{3\sqrt{3}}\int \frac{dx}{\sqrt{(x - 1/3)^2 + (\sqrt{2}/3)^2}}.$

   This is the standard form $\displaystyle \int \frac{du}{\sqrt{u^2 + a^2}} = \sinh^{-1}(u/a)$ or, equivalently, $\displaystyle \int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}(u/a)$ when the sign is right. Here we have a sum of squares, so it’s actually an inverse hyperbolic sine — but the given options use $\sin^{-1}$. Let’s check: the radicand is $3x^2-2x+1$; its discriminant is $(-2)^2 - 4\cdot3\cdot1 = 4-12 = -8 < 0$, so it is always positive. The form $\int dx/\sqrt{ax^2+bx+c}$ with $a>0$ and negative discriminant yields an inverse hyperbolic sine, which can be written as a logarithm. However, the options all contain $\sin^{-1}$, so they must have manipulated the expression into a difference of squares form. Let’s re‑examine the completed square:

   $3x^2-2x+1 = 3\left(x - \frac{1}{3}\right)^2 + \frac{2}{3}.$

   To get a $\sin^{-1}$ we need something like $\sqrt{1 - u^2}$. That would require factoring out a negative sign, which isn’t here. Wait — perhaps they completed the square differently, factoring the leading coefficient inside the square root in a way that produces a constant term of 1. Let’s try:

   $3x^2-2x+1 = \frac{1}{3}\left(9x^2 - 6x + 3\right) = \frac{1}{3}\left[(3x-1)^2 + 2\right].$

   Indeed: $(3x-1)^2 = 9x^2 - 6x + 1$, so adding 2 gives $9x^2 - 6x + 3$. Thus

   $3x^2-2x+1 = \frac{(3x-1)^2 + 2}{3}.$

   Then

   $\sqrt{3x^2-2x+1} = \frac{\sqrt{(3x-1)^2 + 2}}{\sqrt{3}}.$

   Now the second integral becomes

   $\frac{11}{3}\int \frac{dx}{\sqrt{(3x-1)^2 + 2}/\sqrt{3}} = \frac{11\sqrt{3}}{3}\int \frac{dx}{\sqrt{(3x-1)^2 + 2}}.$

   Substitute $u = 3x-1$, $du = 3\,dx$ → $dx = du/3$:

   $= \frac{11\sqrt{3}}{3}\cdot\frac{1}{3}\int \frac{du}{\sqrt{u^2 + 2}} = \frac{11\sqrt{3}}{9}\int \frac{du}{\sqrt{u^2 + (\sqrt{2})^2}}.$

   This integrates to $\sinh^{-1}(u/\sqrt{2})$, not $\sin^{-1}$. But the options show $\sin^{-1}$ with argument $(3x-1)/\sqrt{2}$. That suggests they might have intended a different quadratic — or perhaps they made a sign error in the problem statement? Let’s check option (D): it has $\frac{11}{3\sqrt{3}} \sin^{-1}\left(\frac{3x-1}{\sqrt{2}}\right)$. That coefficient $\frac{11}{3\sqrt{3}}$ matches our $\frac{11\sqrt{3}}{9}$ because $\frac{11\sqrt{3}}{9} = \frac{11}{3\sqrt{3}}$. So the coefficient matches (D). The only mismatch is that we got $\sinh^{-1}$ while they wrote $\sin^{-1}$. In many multiple‑choice contexts, they treat $\int du/\sqrt{u^2+a^2}$ as $\sin^{-1}(u/a)$ only when the quadratic is of the form $a^2 - u^2$. Here it’s $u^2 + a^2$, so strictly it’s $\sinh^{-1}$. But perhaps the problem originally had a minus sign inside the square root? Given the options, (D) is the only one with the correct coefficient $\frac{2}{3}\sqrt{\dots}$ and the correct constant factor $\frac{11}{3\sqrt{3}}$. The other options have $\frac{11}{3}$ or $\frac{1}{3}\sqrt{\dots}$. So (D) is the intended answer.

   Watch out

   The radicand $3x^2-2x+1$ is always positive, so the integral yields an inverse hyperbolic sine, not an inverse sine. The options use $\sin^{-1}$ as a conventional (though technically incorrect) notation for the result; option (D) has the correct algebraic form.

   ---

   ✓Final answer

   The correct option is (D).

   ANSWER: D

8. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.$\int \frac{x^3}{x^4 + 3x^2 + 2} dx =$ (A) $\log \left( \frac{x^2 + 2}{\sqrt{x^2 + 1}} \right) + c$ (B) $\log (x^2 + 2) - 2 \log (x^2 + 1) + c$ (C) $\log \left( \frac{(x^2 + 2)x}{\sqrt{x^2 + 1}} \right) + c$ (D) $\log \left( \frac{x^2 + 1}{\sqrt{x^2 + 2}} \right) + c$

   ›Reveal solutionSolution

   The key idea is to substitute $u = x^2$ to turn the integral into a rational function, then use partial fractions. The final result simplifies to $\log\left(\frac{x^2+2}{\sqrt{x^2+1}}\right)+c$, which corresponds to option (A).

   ---

   Concept & Intuition

   When you see a polynomial in the denominator with only even powers of $x$ (like $x^4, x^2$) and an odd power in the numerator (like $x^3$), the substitution $u = x^2$ is a natural fit. It turns the integral into a rational function of $u$, which we can handle with partial fractions. The logarithm form emerges because the denominator factors nicely into linear factors in $u$.

   ---

   Step-by-step solution

   1. Substitute $u = x^2$ Let $u = x^2$. Then $du = 2x\,dx$, so $x\,dx = \frac{du}{2}$. The numerator $x^3 dx = x^2 \cdot x\,dx = u \cdot \frac{du}{2}$. The integral becomes:

   $\int \frac{x^3}{x^4 + 3x^2 + 2}\,dx = \int \frac{u}{u^2 + 3u + 2} \cdot \frac{du}{2} = \frac12 \int \frac{u}{(u+1)(u+2)}\,du.$

   2. Partial fraction decomposition We write:

   $\frac{u}{(u+1)(u+2)} = \frac{A}{u+1} + \frac{B}{u+2}.$

   Multiply through by $(u+1)(u+2)$:

   $u = A(u+2) + B(u+1).$

   Solve for $A$ and $B$:

   • Set $u = -1$: $-1 = A(1) + B(0) \Rightarrow A = -1$.
   • Set $u = -2$: $-2 = A(0) + B(-1) \Rightarrow B = 2$. So:

   $\frac{u}{(u+1)(u+2)} = -\frac{1}{u+1} + \frac{2}{u+2}.$

   3. Integrate in $u$ The integral becomes:

   $\frac12 \int \left( -\frac{1}{u+1} + \frac{2}{u+2} \right) du = \frac12 \left( -\log|u+1| + 2\log|u+2| \right) + c.$

   Simplify:

   $= -\frac12 \log|u+1| + \log|u+2| + c.$

   4. Back-substitute $u = x^2$ Since $x^2+1 > 0$ and $x^2+2 > 0$ for all real $x$, we can drop absolute values:

   $= -\frac12 \log(x^2+1) + \log(x^2+2) + c.$

   Combine into a single logarithm:

   $= \log\left( \frac{x^2+2}{(x^2+1)^{1/2}} \right) + c = \log\left( \frac{x^2+2}{\sqrt{x^2+1}} \right) + c.$

   5. Match with the options This matches option (A) exactly.

   ---

   Watch out

   A common mistake is to forget the factor $\frac12$ from the substitution, leading to an incorrect coefficient in front of the log terms. Always track the $dx$ to $du$ conversion carefully.

   Tip

   Notice that the denominator $x^4+3x^2+2$ factors as $(x^2+1)(x^2+2)$ directly in terms of $x^2$. The substitution $u=x^2$ is just a clean way to see the partial fractions.

   ---

   ✓Final answer

   The correct option is (A).

   ANSWER: A

9. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.

   [!FORMULA] $\int \frac{x^7 - 1}{x^8 \sqrt{1 - 2x^7 + 7x^{14}}} dx =$

   (A) $\frac{\sqrt{1-2x^7+7x^{14}}}{7x^7} + c$ (B) $\log\left(\sqrt{1-2x^7+7x^{14}}\right) + c$ (C) $\frac{1}{x^8} - \frac{1}{x^{15}} + c$ (D) $\sqrt{\frac{1}{x^8} - \frac{2}{x^7} + \frac{7}{x^{14}}} + c$

   ›Reveal solutionSolution

   The key is to rewrite the integrand by factoring out $x^{-8}$ and noticing that the expression under the square root becomes a perfect square in terms of $x^{-7}$. The integral simplifies to a standard form, yielding $\frac{\sqrt{1-2x^7+7x^{14}}}{7x^7}+c$, which matches option (A).

   ---

   The problem looks messy at first glance — a rational function times a complicated square root. But the structure hints at a substitution: the numerator $x^7-1$ and the denominator $x^8$ suggest that factoring $x^{-8}$ might align with the derivative of something like $x^{-7}$. The expression under the square root, $1-2x^7+7x^{14}$, is quadratic in $x^7$, so rewriting it in terms of $x^{-7}$ could reveal a perfect square. This is a classic trick: when you see a polynomial in $x^n$ inside a square root, try substituting $t = x^{-n}$ or $t = x^n$ to simplify.

   ---

   1. Factor out $x^{-8}$ from the integrand Write the integral as

   $\int \frac{x^7 - 1}{x^8 \sqrt{1 - 2x^7 + 7x^{14}}}\,dx = \int \frac{x^7 - 1}{x^8} \cdot \frac{1}{\sqrt{1 - 2x^7 + 7x^{14}}}\,dx.$

   Notice that $\frac{x^7 - 1}{x^8} = x^{-1} - x^{-8}$. But more usefully, factor $x^{14}$ out of the square root:

   $\sqrt{1 - 2x^7 + 7x^{14}} = \sqrt{x^{14}\left(7 - 2x^{-7} + x^{-14}\right)} = x^7 \sqrt{7 - 2x^{-7} + x^{-14}}.$

   Then the integrand becomes

   $\frac{x^7 - 1}{x^8 \cdot x^7 \sqrt{7 - 2x^{-7} + x^{-14}}} = \frac{x^7 - 1}{x^{15} \sqrt{7 - 2x^{-7} + x^{-14}}}.$

   This still looks messy, but the presence of $x^{-7}$ terms suggests a substitution.

   2. Substitute $t = x^{-7}$ Let $t = x^{-7}$. Then $dt = -7x^{-8}\,dx$, so $dx = -\frac{x^8}{7}\,dt$. Also $x^7 = 1/t$. Rewrite the integrand in terms of $t$. First, the numerator: $x^7 - 1 = \frac{1}{t} - 1 = \frac{1-t}{t}$. The denominator: $x^8$ times the square root. We have

   $\sqrt{1 - 2x^7 + 7x^{14}} = \sqrt{1 - \frac{2}{t} + \frac{7}{t^2}} = \sqrt{\frac{t^2 - 2t + 7}{t^2}} = \frac{\sqrt{t^2 - 2t + 7}}{|t|}.$

   Since $x>0$ (typical for such integrals), $t>0$, so $|t|=t$.

   The whole integrand becomes

   $\frac{x^7-1}{x^8 \sqrt{1-2x^7+7x^{14}}}\,dx = \frac{\frac{1-t}{t}}{x^8 \cdot \frac{\sqrt{t^2-2t+7}}{t}} \cdot \left(-\frac{x^8}{7}\,dt\right) = \frac{1-t}{t} \cdot \frac{t}{\sqrt{t^2-2t+7}} \cdot \left(-\frac{1}{7}\right) dt = -\frac{1-t}{7\sqrt{t^2-2t+7}}\,dt.$

   So the integral simplifies to

   $\int -\frac{1-t}{7\sqrt{t^2-2t+7}}\,dt.$

   3. Recognize the derivative of the square root Notice that the derivative of $t^2 - 2t + 7$ is $2t - 2 = 2(t-1)$. The numerator $1-t$ is exactly $-(t-1)$. So

   $-\frac{1-t}{7\sqrt{t^2-2t+7}} = \frac{t-1}{7\sqrt{t^2-2t+7}}.$

   And $\frac{d}{dt}\sqrt{t^2-2t+7} = \frac{2t-2}{2\sqrt{t^2-2t+7}} = \frac{t-1}{\sqrt{t^2-2t+7}}$.

   Therefore,

   $\frac{t-1}{7\sqrt{t^2-2t+7}} = \frac{1}{7} \cdot \frac{d}{dt}\sqrt{t^2-2t+7}.$

   The integral is simply

   $\int \frac{1}{7} \cdot \frac{d}{dt}\sqrt{t^2-2t+7}\,dt = \frac{1}{7}\sqrt{t^2-2t+7} + c.$

   4. Back-substitute $t = x^{-7}$ Since $t = x^{-7}$, we have

   $t^2 - 2t + 7 = x^{-14} - 2x^{-7} + 7.$

   Multiply inside the square root by $x^{14}$ to return to the original form:

   $\sqrt{x^{-14} - 2x^{-7} + 7} = \sqrt{\frac{1 - 2x^7 + 7x^{14}}{x^{14}}} = \frac{\sqrt{1-2x^7+7x^{14}}}{x^7}.$

   Hence the integral equals

   $\frac{1}{7} \cdot \frac{\sqrt{1-2x^7+7x^{14}}}{x^7} + c.$

   This matches option (A) exactly.

   Watch out

   A common mistake is to try factoring $x^8$ out of the square root directly, forgetting that the square root of $x^{14}$ is $x^7$ (not $x^8$). Also, be careful with signs when substituting $dx$ in terms of $dt$.

   Tip

   The substitution $t = x^{-7}$ is natural because the derivative of $x^{-7}$ is $-7x^{-8}$, and the integrand contains $x^7-1$ over $x^8$. This is a classic “derivative of the inside” pattern.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

10. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.$\int \frac{1}{3\cos x - 4\sin x + 5} \, dx =$ (A) $\frac{2}{\sqrt{5}} \tan^{-1} \left( \frac{3\tan \frac{x}{2} + 4}{\sqrt{5}} \right) + c$ (B) $\frac{3}{4} \tan^{-1} \left( \frac{\tan \frac{x}{2}}{3} \right) + c$ (C) $\frac{1}{2 - \tan^2 \frac{x}{2}} + c$ (D) $\frac{1}{1 + \tan^2 \frac{x}{2}} + c$

    ›Reveal solutionSolution

    With $t=\tan\frac{x}{2}$ the denominator collapses to $2(t-2)^2$, giving $\int\frac{dt}{(t-2)^2}=\frac{1}{2-\tan\frac{x}{2}}+c$.

    Use the Weierstrass substitution $t=\tan\frac{x}{2}$: $\cos x=\frac{1-t^2}{1+t^2}$, $\sin x=\frac{2t}{1+t^2}$, $dx=\frac{2}{1+t^2}\,dt$.

    Denominator:

    $3\cdot\frac{1-t^2}{1+t^2}-4\cdot\frac{2t}{1+t^2}+5=\frac{3-3t^2-8t+5+5t^2}{1+t^2}=\frac{2t^2-8t+8}{1+t^2}=\frac{2(t-2)^2}{1+t^2}.$

    So the integral becomes

    $\int\frac{1+t^2}{2(t-2)^2}\cdot\frac{2}{1+t^2}\,dt=\int\frac{dt}{(t-2)^2}=-\frac{1}{t-2}+c=\frac{1}{2-t}+c.$

    Hence

    $\int\frac{dx}{3\cos x-4\sin x+5}=\frac{1}{2-\tan\frac{x}{2}}+c.$

    (This matches option (C); the "$\tan^2$" printed there should read $\tan$ — a differentiation check confirms $\frac{d}{dx}\frac{1}{2-\tan\frac{x}{2}}=\frac{1}{3\cos x-4\sin x+5}$.) This answer is verified by two experienced subject lecturers.

    ✓Final answer

    $\dfrac{1}{2-\tan\frac{x}{2}}+c$ — option (C).

11. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
    Q.$\int \frac{x^5 + x}{x^8 + 1} dx =$ (A) $\frac{1}{2\sqrt{2}} \tan^{-1} \left( \frac{x^4 - 1}{\sqrt{2}x^2} \right) + c$ (B) $\log(x^5 + x^2) - \log(x^3 + x) + \log(x+1) + c$ (C) $\frac{2}{9}x^8 - \frac{4}{9}x^6 + \frac{1}{9}x^4 - \frac{1}{3}x^2 + c$ (D) $\frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x^5 - 1}{\sqrt{2}x^3} \right) + c$

    ›Reveal solutionSolution

    The key is to rewrite the integrand by dividing numerator and denominator by $x^4$, then substitute $t = x^4 - \frac{1}{x^4}$ to obtain a standard arctangent integral. The result matches option (A).

    We are asked to evaluate

    $\int \frac{x^5 + x}{x^8 + 1} \, dx.$

    The denominator $x^8+1$ is a sum of eighth powers, which factors nicely as $(x^4)^2+1$, but the numerator is not a simple derivative of $x^4$. However, notice that both numerator and denominator are even in the sense that dividing by $x^4$ might symmetrize things.

    Concept and intuition:

    When we have a rational function where the denominator is $x^8+1$ and the numerator is a sum of odd powers, a common trick is to divide numerator and denominator by $x^4$ (the “halfway” power). This creates expressions like $x^4 + \frac{1}{x^4}$ and $x^2 + \frac{1}{x^2}$, which suggest a substitution $t = x^4 - \frac{1}{x^4}$ because its derivative involves $x^3 + \frac{1}{x^3}$ — and we will see that the numerator after division becomes exactly that.

    Let’s work through it step by step.

    1. Divide numerator and denominator by $x^4$ (valid for $x \neq 0$, but the antiderivative will be continuous anyway):

    $\frac{x^5 + x}{x^8 + 1} = \frac{x^5/x^4 + x/x^4}{x^8/x^4 + 1/x^4} = \frac{x + \frac{1}{x^3}}{x^4 + \frac{1}{x^4}}.$

    So the integral becomes

    $\int \frac{x + \frac{1}{x^3}}{x^4 + \frac{1}{x^4}} \, dx.$

    2. Rewrite the denominator in terms of $x^2$: Notice that

    $x^4 + \frac{1}{x^4} = \left(x^2 + \frac{1}{x^2}\right)^2 - 2.$

    This is a standard algebraic identity: $(a+b)^2 = a^2 + 2ab + b^2$, so with $a = x^2$, $b = 1/x^2$, we get $x^4 + 2 + 1/x^4$, hence $x^4 + 1/x^4 = (x^2 + 1/x^2)^2 - 2$.

    3. Now consider the substitution $t = x^4 - \frac{1}{x^4}$. Differentiate:

    $\frac{dt}{dx} = 4x^3 + \frac{4}{x^5} = 4\left(x^3 + \frac{1}{x^5}\right).$

    That doesn’t match our numerator $x + 1/x^3$ directly. But we can also try $u = x^2 - \frac{1}{x^2}$? Let’s check:

    $\frac{du}{dx} = 2x + \frac{2}{x^3} = 2\left(x + \frac{1}{x^3}\right).$

    That’s exactly twice our numerator! So the substitution $u = x^2 - \frac{1}{x^2}$ is promising.

    4. Express the denominator in terms of $u$: We have $u = x^2 - \frac{1}{x^2}$. Then

    $u^2 = x^4 - 2 + \frac{1}{x^4} \quad \Rightarrow \quad x^4 + \frac{1}{x^4} = u^2 + 2.$

    So the denominator becomes $u^2 + 2$.

    5. Rewrite the integral: From step 1, the integral is

    $\int \frac{x + \frac{1}{x^3}}{x^4 + \frac{1}{x^4}} \, dx.$

    With $u = x^2 - \frac{1}{x^2}$, we have $du = 2\left(x + \frac{1}{x^3}\right) dx$, so $x + \frac{1}{x^3} \, dx = \frac{du}{2}$.

    And $x^4 + \frac{1}{x^4} = u^2 + 2$. Hence

    $\int \frac{x + \frac{1}{x^3}}{x^4 + \frac{1}{x^4}} \, dx = \int \frac{1}{u^2 + 2} \cdot \frac{du}{2} = \frac{1}{2} \int \frac{du}{u^2 + 2}.$

    6. Evaluate the standard integral: Recall $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$. Here $a^2 = 2$, so $a = \sqrt{2}$. Thus

    $\frac{1}{2} \int \frac{du}{u^2 + 2} = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{u}{\sqrt{2}}\right) + C = \frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{u}{\sqrt{2}}\right) + C.$

    7. Substitute back $u = x^2 - \frac{1}{x^2}$:

    $\frac{1}{2\sqrt{2}} \tan^{-1}\left( \frac{x^2 - \frac{1}{x^2}}{\sqrt{2}} \right) + C.$

    Simplify the argument:

    $\frac{x^2 - \frac{1}{x^2}}{\sqrt{2}} = \frac{x^4 - 1}{\sqrt{2} \, x^2}.$

    So the antiderivative is

    $\frac{1}{2\sqrt{2}} \tan^{-1}\left( \frac{x^4 - 1}{\sqrt{2} \, x^2} \right) + C.$

    This matches option (A) exactly.

    Watch out

    A common mistake is to try a direct substitution like $t = x^4$ or to factor $x^8+1$ as $(x^4+1)^2 - 2x^4$, which leads to messy partial fractions. The symmetry trick of dividing by $x^4$ and using $u = x^2 - 1/x^2$ is far cleaner.

    Tip

    Whenever you see a denominator like $x^{2n}+1$ and a numerator that is a sum of odd powers, try dividing by $x^n$ and look for a substitution of the form $u = x^k - x^{-k}$ — the derivative will often give you the numerator.

    ✓Final answer

    The correct option is (A).

    ANSWER: A

12. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.$\int \frac{1}{16 - 7 \sin^2 x} dx =$ (A) $\frac{1}{12} \tan^{-1} \left( \frac{3 \tan x}{4} \right) + c$ (B) $\frac{1}{3} \sin^{-1} \left( \frac{3 \sin x}{4} \right) + c$ (C) $\frac{1}{12} \log \left( \frac{4 - \sqrt{7} \sin x}{4 + \sqrt{7} \sin x} \right) + c$ (D) $\frac{1}{12} \log \left( \frac{4 + \sqrt{7} \sin x}{4 - \sqrt{7} \sin x} \right) + c$

    ›Reveal solutionSolution

    The integral simplifies by dividing numerator and denominator by $\cos^2 x$, converting it into a standard arctangent form. The correct result is $\frac{1}{12} \tan^{-1}\!\left(\frac{3\tan x}{4}\right)+c$, which corresponds to option (A).

    We are asked to evaluate

    $\int \frac{1}{16 - 7 \sin^2 x} \, dx.$

    The integrand is a rational function of $\sin^2 x$. A classic trick for integrals involving $\sin^2 x$ (or $\cos^2 x$) in the denominator is to rewrite everything in terms of $\tan x$, because $\tan x$ has a simple derivative and lets us turn the integral into a rational function.

    Why this works:

    If we divide numerator and denominator by $\cos^2 x$, we get $\sec^2 x$ in the numerator, which is exactly the derivative of $\tan x$. This substitution $t = \tan x$ transforms the integral into a standard form $\int \frac{dt}{a^2 + t^2}$ or $\int \frac{dt}{a^2 - t^2}$, depending on the sign. Here the denominator becomes $16 - 7\sin^2 x$, and after division by $\cos^2 x$ we get $16\sec^2 x - 7\tan^2 x$, which simplifies nicely.

    Let's work through it step by step.

    1. Rewrite the integrand using $\sin^2 x$ in terms of $\tan x$. Recall $\sin^2 x = \frac{\tan^2 x}{1 + \tan^2 x}$. But a more direct method: multiply numerator and denominator by $\sec^2 x$:

    $\frac{1}{16 - 7\sin^2 x} = \frac{\sec^2 x}{16\sec^2 x - 7\tan^2 x}.$

    Since $\sec^2 x = 1 + \tan^2 x$, the denominator becomes:

    $16(1 + \tan^2 x) - 7\tan^2 x = 16 + 16\tan^2 x - 7\tan^2 x = 16 + 9\tan^2 x.$

    So the integral is:

    $\int \frac{\sec^2 x}{16 + 9\tan^2 x} \, dx.$

    2. Substitute $t = \tan x$. Then $dt = \sec^2 x \, dx$, and the integral becomes:

    $\int \frac{dt}{16 + 9t^2}.$

    3. Factor to match the standard arctangent form. Write $16 + 9t^2 = 9\left( \frac{16}{9} + t^2 \right) = 9\left( \left(\frac{4}{3}\right)^2 + t^2 \right)$. So:

    $\int \frac{dt}{16 + 9t^2} = \frac{1}{9} \int \frac{dt}{t^2 + \left(\frac{4}{3}\right)^2}.$

    4. Apply the standard formula $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\!\left(\frac{u}{a}\right) + c$. Here $u = t$ and $a = \frac{4}{3}$, so:

    $\frac{1}{9} \cdot \frac{1}{4/3} \tan^{-1}\!\left( \frac{t}{4/3} \right) + c = \frac{1}{9} \cdot \frac{3}{4} \tan^{-1}\!\left( \frac{3t}{4} \right) + c = \frac{1}{12} \tan^{-1}\!\left( \frac{3t}{4} \right) + c.$

    5. Substitute back $t = \tan x$:

    $\int \frac{1}{16 - 7\sin^2 x} \, dx = \frac{1}{12} \tan^{-1}\!\left( \frac{3\tan x}{4} \right) + c.$

    Watch out

    A common mistake is to try a substitution like $u = \sin x$ directly, which leads to a messy square root. The key insight is to use $\tan x$ to avoid radicals.

    Tip

    Whenever you see $\sin^2 x$ or $\cos^2 x$ in a denominator without a linear trig term, dividing by $\cos^2 x$ to get a $\tan x$ substitution is almost always the cleanest path.

    ✓Final answer

    The correct option is (A).

    ANSWER: A