Q.Integrate the following function: $\frac{1}{\sqrt{7 - 6x - x^2}}$

Showing the 12 most recent of 20 on this concept.

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

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

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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 $$$$

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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

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

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5. 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 $\frac{1}{\sec \theta} = \cos \theta$:
   $$ \frac{1}{4} \int \cos \theta \, d\theta $$ …
   

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6. 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

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

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7. 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

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

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

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9. 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: …

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10. 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 $u = x^2$ …

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11. 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 $t = \sqrt{x+3}$. …

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12. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.$\int \frac{3x+2}{4x^2+4x+5} dx = A \log \left(4x^2+4x+5\right) + B \tan^{-1} \left(\frac{2x+1}{2}\right) + c$, then $A+B=$ (A) $\frac{1}{2}$ (B) $\frac{3}{4}$ (C) $\frac{3}{8}$ (D) $\frac{1}{8}$

    ›Reveal solutionSolution

    The integral splits into a logarithmic part (from the derivative of the denominator) and an arctangent part (by completing the square). Matching coefficients gives $A = \frac{3}{8}$ and $B = \frac{3}{4}$, so $A+B = \frac{9}{8}$, which is not among the options — but careful: the problem’s given form uses $A$ and $B$ as constants in front of the log and arctan terms, and the sum is $\frac{9}{8}$. However, re-checking the options, the intended answer is $\frac{3}{8}$ for $A$ alone? No — let’s solve properly.

    ---

    Concept & Intuition

    When integrating a rational function where the denominator is a quadratic that doesn’t factor over the reals, the standard strategy is:

    • If the numerator is (a constant times) the derivative of the denominator, the integral is a logarithm.
    • Otherwise, we split the numerator into a part that is a multiple of the derivative (giving log) plus a constant remainder (giving an arctan after completing the square).

    Here the denominator is $4x^2+4x+5$. Its derivative is $8x+4$. Our numerator is $3x+2$. We write $3x+2$ as $\frac{3}{8}(8x+4) + \text{constant}$ to match the derivative.

    ---

    Step-by-step solution

    1. Find the derivative of the denominator Let $D = 4x^2+4x+5$. Then

    $D' = 8x+4.$

    We want to express $3x+2$ in terms of $8x+4$.

    2. Express numerator as a multiple of $D'$ plus a constant Write

    $3x+2 = \alpha (8x+4) + \beta.$

    Comparing coefficients of $x$: $3 = 8\alpha \Rightarrow \alpha = \frac{3}{8}$.

    Comparing constants: $2 = 4\alpha + \beta \Rightarrow 2 = 4\cdot\frac{3}{8} + \beta = \frac{3}{2} + \beta \Rightarrow \beta = 2 - \frac{3}{2} = \frac{1}{2}$.

    So

    $3x+2 = \frac{3}{8}(8x+4) + \frac{1}{2}.$

    3. Split the integral

    $\int \frac{3x+2}{4x^2+4x+5}\,dx = \frac{3}{8} \int \frac{8x+4}{4x^2+4x+5}\,dx + \frac{1}{2} \int \frac{1}{4x^2+4x+5}\,dx.$

    The first integral is $\log|4x^2+4x+5|$ (since denominator is always positive, we drop absolute value).

    So first part = $\frac{3}{8} \log(4x^2+4x+5)$.

    4. Handle the second integral by completing the square

    $4x^2+4x+5 = 4\left(x^2 + x + \frac{5}{4}\right) = 4\left[\left(x+\frac{1}{2}\right)^2 + 1\right].$

    Because $x^2 + x = (x+1/2)^2 - 1/4$, so $x^2+x+5/4 = (x+1/2)^2 + 1$.

    Thus

    $\frac{1}{2} \int \frac{1}{4x^2+4x+5}\,dx = \frac{1}{2} \int \frac{1}{4\left[(x+1/2)^2 + 1\right]}\,dx = \frac{1}{8} \int \frac{1}{(x+1/2)^2 + 1}\,dx.$

    5. Use the arctan formula

    $\int \frac{1}{u^2+1}\,du = \tan^{-1} u.$

    Let $u = x + \frac{1}{2}$, then $du = dx$. So

    $\frac{1}{8} \int \frac{1}{u^2+1}\,du = \frac{1}{8} \tan^{-1}\left(x+\frac{1}{2}\right).$ …

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