Q.Integrate the following function: $\frac{5x-2}{1+2x+3x^2}$

Showing the 12 most recent of 20 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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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 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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8. 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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9. 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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10. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
    Q.Given that $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + c.$ If $\int \frac{1}{x^4 + 3x^2 + 1} dx = a \cdot \tan^{-1} \left( \frac{b(x^2 - 1)}{x} \right) + c \tan^{-1} \left( \frac{d(x^2 + 1)}{x} \right) + k$ where $k$ is a constant of integration, then $5(c + d + ab) =$ (A) 3 (B) 5 (C) 8 (D) 10

    ›Reveal solutionSolution

    The key is to factor the quartic denominator into two quadratics, then decompose the integrand using partial fractions, integrate each term via the given arctan formula, and match coefficients to find $a, b, c, d$; the final value is $5(c + d + ab) = 5$.

    We are given the standard integral

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

    Our task is to evaluate

    $\int \frac{1}{x^4 + 3x^2 + 1}\,dx$

    and express it in the form

    $a \cdot \tan^{-1}\!\left(\frac{b(x^2 - 1)}{x}\right) + c \cdot \tan^{-1}\!\left(\frac{d(x^2 + 1)}{x}\right) + k.$

    Then we must compute $5(c + d + ab)$ and pick the correct option.

    ---

    1. Factor the denominator

    The denominator $x^4 + 3x^2 + 1$ is a quadratic in $x^2$. Let $t = x^2$:

    $t^2 + 3t + 1 = 0 \quad\Rightarrow\quad t = \frac{-3 \pm \sqrt{5}}{2}.$

    These are negative, so the factorization over reals is into two irreducible quadratics. A clever trick: write

    $x^4 + 3x^2 + 1 = (x^2 + ax + 1)(x^2 - ax + 1).$

    Multiplying out:

    $(x^2 + ax + 1)(x^2 - ax + 1) = x^4 + (2 - a^2)x^2 + 1.$

    We need $2 - a^2 = 3$, so $a^2 = -1$? That’s not right — let’s adjust. Actually try

    $(x^2 + px + 1)(x^2 + qx + 1) = x^4 + (p+q)x^3 + (2 + pq)x^2 + (p+q)x + 1.$

    We want no $x^3$ or $x$ terms, so $p+q = 0$. Then $q = -p$. Then the $x^2$ coefficient is $2 - p^2 = 3$, so $p^2 = -1$ — impossible.

    So the correct factorization uses different constant terms. Try

    $x^4 + 3x^2 + 1 = (x^2 + \alpha x + \beta)(x^2 - \alpha x + \beta).$

    Multiplying:

    $(x^2 + \alpha x + \beta)(x^2 - \alpha x + \beta) = x^4 + (2\beta - \alpha^2)x^2 + \beta^2.$

    We need $\beta^2 = 1$ and $2\beta - \alpha^2 = 3$.

    • If $\beta = 1$, then $2 - \alpha^2 = 3 \Rightarrow \alpha^2 = -1$ (no).
    • If $\beta = -1$, then $-2 - \alpha^2 = 3 \Rightarrow \alpha^2 = -5$ (no).

    Thus the factorization must involve irrational coefficients. Solve $x^4 + 3x^2 + 1 = 0$ for $x^2$:

    $x^2 = \frac{-3 \pm \sqrt{5}}{2}.$

    So

    $x^4 + 3x^2 + 1 = \left(x^2 - \frac{-3 + \sqrt{5}}{2}\right)\left(x^2 - \frac{-3 - \sqrt{5}}{2}\right).$

    But these are of the form $(x^2 + p)(x^2 + q)$ with $p,q > 0$. Indeed,

    $p = \frac{3 - \sqrt{5}}{2}, \quad q = \frac{3 + \sqrt{5}}{2}.$

    Check: $p+q = 3$, $pq = 1$. So

    $x^4 + 3x^2 + 1 = (x^2 + p)(x^2 + q), \quad p = \frac{3 - \sqrt{5}}{2},\; q = \frac{3 + \sqrt{5}}{2}.$

    ---

    2. Partial fractions

    We write

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

    Since the numerator is constant and denominators are even, the simplest form is

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

    Check:

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

    Yes. So

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

    Now $q-p = \sqrt{5}$. So

    $= \frac{1}{\sqrt{5}} \left( \frac{1}{\sqrt{p}} \tan^{-1}\!\left(\frac{x}{\sqrt{p}}\right) - \frac{1}{\sqrt{q}} \tan^{-1}\!\left(\frac{x}{\sqrt{q}}\right) \right) + k.$

    ---

    3. Rewrite in the required form

    We have

    $\sqrt{p} = \sqrt{\frac{3 - \sqrt{5}}{2}}, \quad \sqrt{q} = \sqrt{\frac{3 + \sqrt{5}}{2}}.$

    Notice that

    $\sqrt{p} = \frac{\sqrt{5} - 1}{2}, \quad \sqrt{q} = \frac{\sqrt{5} + 1}{2}.$

    Check: $\left(\frac{\sqrt{5}-1}{2}\right)^2 = \frac{5 + 1 - 2\sqrt{5}}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}$. Yes. Similarly for the other.

    Thus

    $\frac{1}{\sqrt{p}} = \frac{2}{\sqrt{5} - 1} = \frac{2(\sqrt{5}+1)}{4} = \frac{\sqrt{5}+1}{2},$

    $\frac{1}{\sqrt{q}} = \frac{2}{\sqrt{5}+1} = \frac{2(\sqrt{5}-1)}{4} = \frac{\sqrt{5}-1}{2}.$

    So the integral becomes

    $\frac{1}{\sqrt{5}} \left( \frac{\sqrt{5}+1}{2} \tan^{-1}\!\left(\frac{2x}{\sqrt{5}-1}\right) - \frac{\sqrt{5}-1}{2} \tan^{-1}\!\left(\frac{2x}{\sqrt{5}+1}\right) \right) + k.$

    Simplify the coefficient:

    $= \frac{1}{2\sqrt{5}} \left( (\sqrt{5}+1) \tan^{-1}\!\left(\frac{2x}{\sqrt{5}-1}\right) - (\sqrt{5}-1) \tan^{-1}\!\left(\frac{2x}{\sqrt{5}+1}\right) \right) + k.$

    ---

    4. Match to the given form

    The given form is

    $a \tan^{-1}\!\left(\frac{b(x^2 - 1)}{x}\right) + c \tan^{-1}\!\left(\frac{d(x^2 + 1)}{x}\right) + k.$

    We need to express our arctan arguments in terms of $(x^2 \pm 1)/x$. Notice:

    $\frac{2x}{\sqrt{5}-1} = \frac{2x(\sqrt{5}+1)}{4} = \frac{(\sqrt{5}+1)x}{2}.$

    That doesn’t look like $(x^2-1)/x$. Instead, use the tangent addition formula trick.

    Recall:

    $\tan^{-1} u + \tan^{-1} v = \tan^{-1}\!\left(\frac{u+v}{1-uv}\right).$

    We suspect that the two arctan terms combine into something like $\tan^{-1}\!\left(\frac{x^2-1}{x}\right)$ and $\tan^{-1}\!\left(\frac{x^2+1}{x}\right)$.

    Let’s set

    $\tan^{-1}\!\left(\frac{2x}{\sqrt{5}-1}\right) = \tan^{-1}\!\left(\frac{x^2-1}{x}\right) + \tan^{-1}\!\left(\frac{1}{\phi}\right)$

    or similar. A systematic approach:

    We know

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

    Also

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

    Now use the identity:

    $\tan^{-1} u - \tan^{-1} v = \tan^{-1}\!\left(\frac{u-v}{1+uv}\right).$

    Compute

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

    That’s not our form.

    Instead, try:

    $\tan^{-1}\!\left(\frac{2x}{\sqrt{5}-1}\right) = \tan^{-1}\!\left( \frac{x^2-1}{x} \cdot \frac{?}{?} \right).$

    Observe that

    $\frac{2}{\sqrt{5}-1} = \frac{\sqrt{5}+1}{2} = \phi \approx 1.618.$

    So the argument is $\phi x$. Meanwhile,

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

    These are different.

    Better: Use the known identity:

    $\tan^{-1}\!\left(\frac{x}{\sqrt{p}}\right) = \frac{1}{2} \tan^{-1}\!\left( \frac{2x\sqrt{p}}{p - x^2} \right) \quad \text{(not helpful)}.$

    Instead, note that the required form has arguments $(x^2-1)/x$ and $(x^2+1)/x$. Let’s set

    $u = \frac{x^2-1}{x}, \quad v = \frac{x^2+1}{x}.$

    Then

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

    Not simple.

    ---

    5. Direct coefficient matching via tangent addition

    We want to write

    $\frac{1}{\sqrt{5}} \left( \frac{1}{\sqrt{p}} \tan^{-1}\!\left(\frac{x}{\sqrt{p}}\right) - \frac{1}{\sqrt{q}} \tan^{-1}\!\left(\frac{x}{\sqrt{q}}\right) \right)$

    as

    $a \tan^{-1}\!\left( \frac{b(x^2-1)}{x} \right) + c \tan^{-1}\!\left( \frac{d(x^2+1)}{x} \right).$

    Differentiate both sides and compare. But that’s messy.

    Instead, use the known result:

    $\tan^{-1}\!\left(\frac{x}{\sqrt{p}}\right) = \frac{1}{2} \tan^{-1}\!\left( \frac{2x\sqrt{p}}{p - x^2} \right) + \text{constant}.$

    But $p = \frac{3-\sqrt{5}}{2}$, so $\sqrt{p} = \frac{\sqrt{5}-1}{2}$. Then

    $\frac{2\sqrt{p}}{p - x^2} = \frac{\sqrt{5}-1}{\frac{3-\sqrt{5}}{2} - x^2} = \frac{2(\sqrt{5}-1)}{3-\sqrt{5} - 2x^2}.$

    Not matching.

    ---

    6. A smarter approach: Use the given form as a hint

    The answer choices are small integers, so $a, b, c, d$ are likely simple. Try to guess:

    Suppose $b = 1$ and $d = 1$. Then the integral is

    $a \tan^{-1}\!\left(\frac{x^2-1}{x}\right) + c \tan^{-1}\!\left(\frac{x^2+1}{x}\right).$

    Differentiate:

    $\frac{d}{dx} \tan^{-1}\!\left(\frac{x^2-1}{x}\right) = \frac{1}{1 + \left(\frac{x^2-1}{x}\right)^2} \cdot \frac{d}{dx}\left( x - \frac{1}{x} \right) = \frac{x^2}{x^4 + 1} \cdot \left(1 + \frac{1}{x^2}\right) = \frac{x^2+1}{x^4+1}.$

    Similarly,

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

    So the derivative of the proposed form is

    $a \cdot \frac{x^2+1}{x^4+1} + c \cdot \frac{x^2-1}{x^4+1} = \frac{(a+c)x^2 + (a-c)}{x^4+1}.$

    But our integrand is $\frac{1}{x^4+3x^2+1}$, not $\frac{1}{x^4+1}$. So this guess fails.

    ---

    7. Try scaling: Let $b$ and $d$ be such that the denominators become $x^4+3x^2+1$.

    We need

    $$$$

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

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