Q.Integrate the following functions w.r.t. $x$:

Showing the 12 most recent of 32 on this concept.

1. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.$\int \dfrac{\sin^{-1} x}{\sqrt{1 - x^2}}\, dx =$ (A) $\dfrac{1}{2}(\sin^{-1} x)^2 + C$ (B) $-(\sin^{-1} x)\sqrt{1 - x^2} + C$ (C) $(\sin^{-1} x)\sqrt{1 - x^2} + x + C$ (D) $(\sin^{-1} x)\sqrt{1 - x^2} - x + C$ (E) $(\sin^{-1} x)^2 + C$

   ›Reveal solutionSolution

   Substituting u=sin^{-1}x turns the integral into u du = (1/2)(sin^{-1}x)^2 + C.

   Concept and Intuition

   The factor 1/sqrt(1-x^2) is exactly the derivative of sin^{-1}x, so the substitution u = sin^{-1}x collapses the integral.

   Step-by-Step Solution

   1. Let u = sin^{-1}x, then du = dx/sqrt(1-x^2).
   2. The integral becomes integral of u du.
   3. That equals u^2/2 + C = (1/2)(sin^{-1}x)^2 + C.

   Common Mistakes

   • Forgetting the factor 1/2 from integrating u.

   ✓Final answer

   The correct option is (A) — (1/2)(sin^{-1}x)^2 + C.

   ANSWER: A

2. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.$\int \frac{\sin(\cot^{-1}x)}{1 + x^2} dx$ is equal to (A) $-\cos(\cot^{-1}x) + C$ (B) $\cos(\cot^{-1}x) + C$ (C) $\frac{\cos(\cot^{-1}x)}{1 + x^2} + C$ (D) $\frac{\cos(\cot^{-1}x)}{2} + C$ (E) $\frac{-\cos(\cot^{-1}x)}{1 + x^2} + C$

   ›Reveal solutionSolution

   Substitute $u=\cot^{-1}x$ so $du=-\tfrac{dx}{1+x^2}$.

   Let $u=\cot^{-1}x\Rightarrow du=-\dfrac{dx}{1+x^2}$, i.e. $\dfrac{dx}{1+x^2}=-du$.

   $\int\frac{\sin(\cot^{-1}x)}{1+x^2}\,dx=\int\sin u\,(-du)=-\left(-\cos u\right)+C=\cos u+C.$

   Back-substituting, $=\cos(\cot^{-1}x)+C$.

   ✓Final answer

   The correct option is (B).

3. KEAM 2025Set eng-2025-04294 marksMCQ
   Q.$\int \dfrac{dx}{\cos x \sqrt{2 \sin 2x}} =$ (A) $\dfrac{1}{2}\sqrt{\tan x} + C$ (B) $\sqrt{\tan x} + C$ (C) $2\sqrt{\tan x} + C$ (D) $4\sqrt{\tan x} + C$ (E) $3\sqrt{\tan x} + C$

   ›Reveal solutionSolution

   Writing $2\sin2x=4\sin x\cos x$ reduces the integrand to $\frac12\sec^2x\,(\tan x)^{-1/2}$; with $u=\tan x$ this integrates to $\sqrt{\tan x}+C$.

   Use $\sin 2x=2\sin x\cos x$, so $2\sin 2x=4\sin x\cos x$ and $\sqrt{2\sin 2x}=2\sqrt{\sin x\cos x}$. Then

   $\frac{1}{\cos x\sqrt{2\sin 2x}}=\frac{1}{2\cos x\sqrt{\sin x\cos x}}=\frac{1}{2\cos^2 x\sqrt{\tan x}}=\frac{\sec^2 x}{2\sqrt{\tan x}}.$

   Let $u=\tan x$, $du=\sec^2 x\,dx$:

   $\frac12\int u^{-1/2}\,du=\frac12\cdot2u^{1/2}=\sqrt{u}=\sqrt{\tan x}+C.$

   ✓Final answer

   The correct option is (B).

4. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.$\displaystyle\int \dfrac{\sqrt{\sqrt{x} + 1}}{\sqrt{x}}\, dx =$ (A) $\dfrac{4}{3}(\sqrt{x} + 1)^{\frac{3}{2}} + C$ (B) $\dfrac{2}{3}(\sqrt{x} + 1)^{\frac{3}{2}} + C$ (C) $\dfrac{4}{3}(\sqrt{x} + 1)^{\frac{3}{4}} + C$ (D) $\dfrac{1}{3}(\sqrt{x} + 1)^{\frac{3}{2}} + C$ (E) $\dfrac{3}{4}(\sqrt{x} + 1)^{\frac{3}{2}} + C$

   ›Reveal solutionSolution

   Substitute $u=\sqrt{x}+1$.

   With $u=\sqrt{x}+1$, $du=\dfrac{1}{2\sqrt{x}}\,dx$, i.e. $\dfrac{dx}{\sqrt{x}}=2\,du$. Then

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

   ✓Final answer

   The correct option is (A).

5. KEAM 2024Set eng-2024-06064 marksMCQ
   Q.$\int\frac{x^{2}-1}{x^{4}+3x^{2}+1}\,dx =$ (A) $\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x^{2}+1}{\sqrt{3}x}\right)+C$ (B) $\tan^{-1}(x^{2}-1)+C$ (C) $\tan^{-1}\left(x-\frac{1}{x}\right)+C$ (D) $\frac{1}{\sqrt{5}}\tan^{-1}\left(\frac{x^{2}+1}{\sqrt{5}x}\right)+C$ (E) $\tan^{-1}\left(x+\frac{1}{x}\right)+C$

   ›Reveal solutionSolution

   The substitution $u=x+\frac{1}{x}$ turns it into $\int\frac{du}{u^2+1}=\tan^{-1}\left(x+\frac{1}{x}\right)+C$.

   Divide numerator and denominator by $x^2$:

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

   Let $u=x+\frac{1}{x}$, so $du=\left(1-\frac{1}{x^2}\right)dx$ and $x^2+\frac{1}{x^2}=u^2-2$. The denominator becomes $u^2-2+3=u^2+1$.

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

   ✓Final answer

   The correct option is (E).

6. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.$\displaystyle\int \dfrac{\sec x\sqrt{\sec x}}{\sqrt{\sin x} + \sqrt{\cos x}}\, dx =$ (A) $2\left(\sqrt{\tan x} - \log_e\left(\sqrt{\tan x} + 1\right)\right) + C$ (B) $2\left(\sqrt{\tan x} + \log_e\left(\sqrt{\tan x} + 1\right)\right) + C$ (C) $2\sqrt{\tan x} - \log_e\left(\sqrt{\tan x} + 1\right) + C$ (D) $2\sqrt{\tan x} + \log_e\left(\sqrt{\tan x} + 1\right) + C$ (E) $\sqrt{\tan x} - \log_e\left(\sqrt{\tan x} + 1\right) + C$

   ›Reveal solutionSolution

   Factor $\sqrt{\sin x}+\sqrt{\cos x}=\sqrt{\cos x}(\sqrt{\tan x}+1)$ to reduce the integrand to $\dfrac{\sec^2 x}{\sqrt{\tan x}+1}$; substituting $u=\sqrt{\tan x}$ gives option (A).

   Write the denominator as $\sqrt{\sin x}+\sqrt{\cos x}=\sqrt{\cos x}\bigl(\sqrt{\tan x}+1\bigr)$. The numerator is $\sec x\sqrt{\sec x}=\cos^{-3/2}x$, so

   $\dfrac{\cos^{-3/2}x}{\sqrt{\cos x}\,(\sqrt{\tan x}+1)}=\dfrac{\cos^{-2}x}{\sqrt{\tan x}+1}=\dfrac{\sec^2 x}{\sqrt{\tan x}+1}.$

   Let $t=\tan x$, $dt=\sec^2 x\,dx$, then $u=\sqrt t$, $t=u^2$, $dt=2u\,du$:

   $\int\dfrac{dt}{\sqrt t+1}=\int\dfrac{2u\,du}{u+1}=2\int\left(1-\dfrac{1}{u+1}\right)du=2\bigl(u-\log(u+1)\bigr)+C.$

   Thus the integral is $2\bigl(\sqrt{\tan x}-\log(\sqrt{\tan x}+1)\bigr)+C$.

   ✓Final answer

   The correct option is (A).

7. KEAM 2024Set eng-2024-06054 marksMCQ
   Q.$\int \frac{\tan^5 x \; \sec^2 x}{\tan^{12} x + 1}\,dx$ is equal to (A) $\frac{1}{6}\tan^{-1}\left[\tan^6 x\right] + C$ (B) $\frac{1}{2}\tan^{-1}\left[\tan^6 x\right] + C$ (C) $\frac{1}{4}\tan^{-1}\left[\tan^4 x\right] + C$ (D) $\frac{1}{3}\tan^{-1}\left[\tan^3 x\right] + C$ (E) $\frac{1}{7}\tan^{-1}\left[\tan^7 x\right] + C$

   ›Reveal solutionSolution

   Substitute $u=\tan^6 x$; the integral reduces to $\tfrac16\int \frac{du}{u^2+1} = \tfrac16\tan^{-1}(\tan^6 x)+C$.

   Let $u = \tan^6 x$. Then $du = 6\tan^5 x \sec^2 x\,dx$, so $\tan^5 x\sec^2 x\,dx = \tfrac{du}{6}$.

   Also $\tan^{12}x = (\tan^6 x)^2 = u^2$.

   $\int \dfrac{\tan^5 x\sec^2 x}{\tan^{12}x+1}\,dx = \dfrac{1}{6}\int \dfrac{du}{u^2+1} = \dfrac{1}{6}\tan^{-1}u + C = \dfrac{1}{6}\tan^{-1}(\tan^6 x)+C$.

   ✓Final answer

   The correct option is (A).

8. KEAM 2024Set eng-2024-06094 marksMCQ
   Q.$\int \dfrac{dx}{x^2 (x^4 + 1)^{3/4}} =$ (A) $-(x^4 + 1)^{1/4} + C$ (B) $(x^4 + 1)^{1/4} + C$ (C) $-\left(\dfrac{x^4 + 1}{x^4}\right)^{1/4} + C$ (D) $\left(\dfrac{x^4 + 1}{x^4}\right) + C$ (E) $\left(\dfrac{x^4 + 1}{x^4}\right)^{3/4} + C$

   ›Reveal solutionSolution

   Factor $x^4$ out of the radical; with $u=1+x^{-4}$, $du=-4x^{-5}dx$, the integral becomes $-u^{1/4}=-\left(\frac{x^4+1}{x^4}\right)^{1/4}+C$.

   Write $(x^4+1)^{3/4}=x^3(1+x^{-4})^{3/4}$, so

   $\int\frac{dx}{x^2(x^4+1)^{3/4}}=\int\frac{dx}{x^5(1+x^{-4})^{3/4}}.$

   Let $u=1+x^{-4}$, then $du=-4x^{-5}\,dx$, i.e. $x^{-5}dx=-\tfrac14\,du$:

   $\int-\tfrac14 u^{-3/4}\,du=-u^{1/4}+C=-(1+x^{-4})^{1/4}+C=-\left(\frac{x^4+1}{x^4}\right)^{1/4}+C.$

   ✓Final answer

   The correct option is (C).

9. KEAM 2021Set eng-2021-P2-B14 marksMCQ
   Q.$\int \dfrac{\cos(\tan x)}{\cos^2 x} \, dx =$ (A) $(\tan x)\sin(\tan x) + C$ (B) $\sin(\tan x) + C$ (C) $\sec(\tan x) + C$ (D) $(\cos x)\sin(\tan x) + C$ (E) $\cos^2(\tan x) + C$

   ›Reveal solutionSolution

   $\int\dfrac{\cos(\tan x)}{\cos^2 x}\,dx=\sin(\tan x)+C$.

   Concept and Intuition

   The factor $1/\cos^2 x=\sec^2 x$ is exactly the derivative of $\tan x$, so substituting $u=\tan x$ linearizes the integral.

   Step-by-Step Solution

   1. Rewrite as $\int\cos(\tan x)\sec^2 x\,dx$.
   2. Let $u=\tan x$, $du=\sec^2 x\,dx$.
   3. Integral $=\int\cos u\,du=\sin u+C=\sin(\tan x)+C$.

   Common Mistakes

   • Failing to recognize $1/\cos^2 x=\sec^2 x$ as $d(\tan x)$.

   ✓Final answer

   The correct option is (B) — $\sin(\tan x) + C$.

   ANSWER: B

10. KEAM 2024Set eng-2024-06064 marksMCQ
    Q.$\int\frac{4x\cos\sqrt{4x^{2}+7}}{\sqrt{4x^{2}+7}}\,dx =$ (A) $\frac{1}{2}\sin\sqrt{4x^{2}+7}+C$ (B) $\frac{7}{2}\sin\sqrt{4x^{2}+7}+C$ (C) $\sin\sqrt{4x^{2}+7}+C$ (D) $\frac{1}{4}\sin\sqrt{4x^{2}+7}+C$ (E) $\frac{7}{4}\sin\sqrt{4x^{2}+7}+C$

    ›Reveal solutionSolution

    With $u=\sqrt{4x^2+7}$ the integrand is exactly $\cos u\,du$, giving $\sin\sqrt{4x^2+7}+C$.

    Let $u=\sqrt{4x^2+7}$. Then $\frac{du}{dx}=\frac{8x}{2\sqrt{4x^2+7}}=\frac{4x}{\sqrt{4x^2+7}}$, so $du=\frac{4x}{\sqrt{4x^2+7}}\,dx$.

    The integral becomes $\int\cos u\,du=\sin u+C=\sin\sqrt{4x^2+7}+C$.

    ✓Final answer

    The correct option is (C).

11. KEAM 2023Set eng-2023-P2-B24 marksMCQ
    Q.$\int \dfrac{2\tan x + 3}{\sin^{2} x + 2\cos^{2} x}\, dx =$ (A) $\dfrac{3}{\sqrt{2}} \sin^{-1}\left(\dfrac{\sin x}{\sqrt{2}}\right) + \ln\left|\sin^{2} x + 2\right| + C$ (B) $\dfrac{3}{\sqrt{2}} \tan^{-1}\left(\dfrac{\tan x}{\sqrt{2}}\right) + \ln\left|\tan^{2} x + 2\right| + C$ (C) $\dfrac{1}{\sqrt{2}} \tan^{-1}\left(\dfrac{\tan x}{\sqrt{2}}\right) - \ln\left|\tan^{2} x + 2\right| + C$ (D) $\dfrac{3}{\sqrt{2}} \cos^{-1}\left(\dfrac{\cos x}{\sqrt{2}}\right) + \ln\left|\sin^{2} x + 2\right| + C$ (E) $\dfrac{1}{\sqrt{2}} \cos^{-1}\left(\dfrac{\cos x}{\sqrt{2}}\right) - \ln\left|\cos^{2} x + 2\right| + C$

    ›Reveal solutionSolution

    The integral is $\dfrac{3}{\sqrt2}\tan^{-1}\!\left(\dfrac{\tan x}{\sqrt2}\right)+\log|\tan^2x+2|+C$.

    Concept and Intuition

    Divide numerator and denominator by $\cos^2x$ to convert everything into $\tan x$, then substitute $t=\tan x$.

    Step-by-Step Solution

    1. Dividing by $\cos^2x$: integrand $=\dfrac{(2\tan x+3)\sec^2x}{\tan^2x+2}$.
    2. Let $t=\tan x,\;dt=\sec^2x\,dx$: $\int\dfrac{2t+3}{t^2+2}\,dt$.
    3. Split: $\int\dfrac{2t}{t^2+2}dt=\log(t^2+2)$ and $\int\dfrac{3}{t^2+2}dt=\dfrac{3}{\sqrt2}\tan^{-1}\dfrac{t}{\sqrt2}$.

    Common Mistakes

    • Not recognising the $2t\,dt$ pattern for the log term.
    • Missing the $1/\sqrt2$ factor in the arctan term.

    ✓Final answer

    The correct option is (B) — $\dfrac{3}{\sqrt2}\tan^{-1}\!\left(\dfrac{\tan x}{\sqrt2}\right)+\log|\tan^2x+2|+C$.

    ANSWER: B

12. KEAM 2022Set eng-2022-P2-B14 marksMCQ
    Q.$\int \frac{\sin^{25} x}{\cos^{27} x} \, dx$ is equal to (A) $\frac{\sin^{26}(x)}{26} + C$ (B) $\frac{\cos^{26}(x)}{26} + C$ (C) $\tan^{26}(x) + C$ (D) $\frac{\tan^{26}(x)}{26} + C$ (E) $26\tan^{26}(x) + C$

    ›Reveal solutionSolution

    The integral equals $\frac{\tan^{26}x}{26}+C$.

    Concept and Intuition

    Split off a $\sec^2x$ factor so the rest becomes a power of $\tan x$, then substitute $u=\tan x$.

    Step-by-Step Solution

    1. $\frac{\sin^{25}x}{\cos^{27}x}=\tan^{25}x\cdot\frac{1}{\cos^2x}=\tan^{25}x\,\sec^2x$.
    2. Let $u=\tan x\Rightarrow du=\sec^2x\,dx$.
    3. $\int \tan^{25}x\,\sec^2x\,dx=\int u^{25}\,du=\frac{u^{26}}{26}+C$.
    4. $=\frac{\tan^{26}x}{26}+C$.

    Common Mistakes

    • Miscounting powers so the $\sec^2$ factor is not isolated.
    • Forgetting the $\frac1{26}$ from integrating $u^{25}$.

    ✓Final answer

    The correct option is (D) — $\frac{\tan^{26}(x)}{26} + C$.

    ANSWER: D