Q.Integrate the following function: $\frac{1}{x+x \log x}$

Showing the 12 most recent of 32 on this concept.

1. 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).

2. KEAM 2025Set eng-2025-04274 marksMCQ
   Q.$\int \dfrac{\cos\theta}{2 - \sin^2\theta}\,d\theta =$ (A) $\dfrac{1}{2}\log\left|\dfrac{\sqrt{2} - \sin\theta}{\sqrt{2} + \sin\theta}\right| + C$ (B) $\dfrac{1}{2}\log\left|\dfrac{\sqrt{2} + \sin\theta}{\sqrt{2} - \sin\theta}\right| + C$ (C) $\log\left|\dfrac{\sqrt{2} + \sin\theta}{\sqrt{2} - \sin\theta}\right| + C$ (D) $\dfrac{1}{\sqrt{2}}\log\left|\dfrac{\sqrt{2} + \sin\theta}{\sqrt{2} - \sin\theta}\right| + C$ (E) $\dfrac{1}{2\sqrt{2}}\log\left|\dfrac{\sqrt{2} + \sin\theta}{\sqrt{2} - \sin\theta}\right| + C$

   ›Reveal solutionSolution

   Substitution $u=\sin\theta$ gives $\int\dfrac{du}{2-u^2}=\dfrac{1}{2\sqrt2}\log\left|\dfrac{\sqrt2+\sin\theta}{\sqrt2-\sin\theta}\right|+C$.

   Let $u=\sin\theta$, so $du=\cos\theta\,d\theta$. The integral becomes

   $\int\frac{du}{2-u^2}=\int\frac{du}{(\sqrt2)^2-u^2}.$

   Using the standard result $\displaystyle\int\frac{du}{a^2-u^2}=\frac{1}{2a}\log\left|\frac{a+u}{a-u}\right|+C$ with $a=\sqrt2$:

   $=\frac{1}{2\sqrt2}\log\left|\frac{\sqrt2+u}{\sqrt2-u}\right|+C=\frac{1}{2\sqrt2}\log\left|\frac{\sqrt2+\sin\theta}{\sqrt2-\sin\theta}\right|+C.$

   ✓Final answer

   The correct option is (E).

3. 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).

4. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.$\int e^x (x^2 - 2)\cos\left(e^x (x^2 - 2x)\right)\, dx =$ (A) $\sin\left(e^x (x^2 - 2x)\right) + C$ (B) $\sin\left(e^x (x^2 - 2)\right) + C$ (C) $x^2 e^x \sin\left(e^x (x^2 - 2)\right) + C$ (D) $e^x \sin\left(e^x (x^2 - 2)\right) + C$ (E) $e^x \sin\left(x^2 e^x - 2x e^x\right) + C$

   ›Reveal solutionSolution

   Since d/dx[e^x(x^2-2x)] = e^x(x^2-2), the integral is sin(e^x(x^2-2x)) + C.

   Concept and Intuition

   Look for the outer function's argument, g(x) = e^x(x^2-2x), and check whether the rest of the integrand is exactly g'(x); if so the integral is a direct substitution.

   Step-by-Step Solution

   1. Let g(x) = e^x(x^2 - 2x).
   2. g'(x) = e^x(x^2-2x) + e^x(2x-2) = e^x(x^2 - 2), which is exactly the prefactor.
   3. So the integrand is cos(g)*g' dx, and integral cos(g) dg = sin(g) + C.
   4. Result: sin(e^x(x^2-2x)) + C.

   Common Mistakes

   • Confusing the argument (x^2-2x) with the prefactor (x^2-2).

   ✓Final answer

   The correct option is (A) — sin(e^x(x^2-2x)) + C.

   ANSWER: A

5. KEAM 2025Set eng-2025-04264 marksMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{1}{1+\sin x}\, dx =$ (A) $2$ (B) $\dfrac{1}{2}$ (C) $\dfrac{1}{4}$ (D) $1$ (E) $0$

   ›Reveal solutionSolution

   Rationalise the denominator, integrate $\sec^2x-\sec x\tan x$.

   Multiply numerator and denominator by $1-\sin x$:

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

   Antiderivative is $\tan x-\sec x=\dfrac{\sin x-1}{\cos x}$. Evaluating from $0$ to $\tfrac{\pi}{2}$: at $x\to\tfrac{\pi}{2}$ the value $\to 0$; at $x=0$ it is $0-1=-1$. Thus the integral $=0-(-1)=1$.

   ✓Final answer

   The correct option is (D).

6. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.$\int \left(\frac{\sin t + \cos t}{13 + 36\sin 2t}\right) dt$ is equal to (A) $\frac{1}{84}\log\left|\frac{7 + 6(\sin t - \cos t)}{7 - 6(\sin t - \cos t)}\right| + C$ (B) $\frac{1}{81}\log\left|\frac{7 + 6(\sin t - \cos t)}{7 - 6(\sin t - \cos t)}\right| + C$ (C) $\frac{1}{84}\log\left|\frac{7 - 6(\sin t - \cos t)}{7 + 6(\sin t - \cos t)}\right| + C$ (D) $\frac{1}{48}\log\left|\frac{7 + 6(\sin t - \cos t)}{7 - 6(\sin t - \cos t)}\right| + C$ (E) $\frac{1}{64}\log\left|\frac{7 + 6(\sin t - \cos t)}{7 - 6(\sin t - \cos t)}\right| + C$

   ›Reveal solutionSolution

   Let $u=\sin t-\cos t$; the numerator becomes $du$ and $13+36\sin 2t=49-36u^2$, giving a standard $\tfrac{du}{a^2-k^2u^2}$ integral.

   Set $u=\sin t-\cos t$. Then

   $du=(\cos t+\sin t)\,dt,$

   which is exactly the numerator, and

   $u^2=1-2\sin t\cos t=1-\sin 2t\;\Rightarrow\;\sin 2t=1-u^2.$

   So the denominator is

   $13+36\sin 2t=13+36(1-u^2)=49-36u^2.$

   Hence

   $\int\frac{du}{49-36u^2}=\int\frac{du}{7^2-(6u)^2}.$

   Using $\int\tfrac{du}{a^2-k^2u^2}=\tfrac{1}{2ak}\log\left|\tfrac{a+ku}{a-ku}\right|$ with $a=7,\;k=6$,

   $=\frac{1}{2\cdot 7\cdot 6}\log\left|\frac{7+6u}{7-6u}\right|=\frac{1}{84}\log\left|\frac{7+6(\sin t-\cos t)}{7-6(\sin t-\cos t)}\right|+C.$

   ✓Final answer

   The correct option is (A).

7. 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).

8. KEAM 2022Set eng-2022-P2-B14 marksMCQ
   Q.$\int \frac{e^{\frac{1}{\sqrt{t}}}}{t\sqrt{t}} \, dt =$ (A) $\frac{1}{2}e^{\frac{1}{\sqrt{t}}} + C$ (B) $\frac{-1}{2}e^{\frac{1}{\sqrt{t}}} + C$ (C) $2e^{\frac{1}{\sqrt{t}}} + C$ (D) $-2e^{\frac{1}{\sqrt{t}}} + C$ (E) $e^{\frac{1}{\sqrt{t}}} + C$

   ›Reveal solutionSolution

   The integral equals $-2e^{1/\sqrt t}+C$.

   Concept and Intuition

   Recognize that the exponent's derivative appears (up to a constant) in the integrand, so a direct substitution works.

   Step-by-Step Solution

   1. Let $u=\frac1{\sqrt t}=t^{-1/2}$.
   2. $du=-\frac12 t^{-3/2}\,dt=-\frac12\cdot\frac{1}{t\sqrt t}\,dt$, so $\frac{dt}{t\sqrt t}=-2\,du$.
   3. $\int \frac{e^{1/\sqrt t}}{t\sqrt t}\,dt=\int e^{u}(-2)\,du=-2e^{u}+C$.
   4. Back-substitute: $=-2e^{1/\sqrt t}+C$.

   Common Mistakes

   • Dropping the factor of $-2$ from the substitution.
   • Sign error in $du$.

   ✓Final answer

   The correct option is (D) — $-2e^{\frac{1}{\sqrt{t}}} + C$.

   ANSWER: D

9. KEAM 2024Set eng-2024-06054 marksMCQ
   Q.$\int \frac{\sec x}{(\sec x + \tan x)^2}\,dx =$ (A) $\frac{2}{5(\sec x + \tan x)^4} + C$ (B) $\frac{-1}{2(\sec x + \tan x)^2} + C$ (C) $\frac{2}{3(\sec x + \tan x)^{3/2}} + C$ (D) $\frac{-2}{3(\sec x + \tan x)^3} + C$ (E) $(\sec x + \tan x)^2 + C$

   ›Reveal solutionSolution

   The substitution $u=\sec x+\tan x$ turns it into $\int u^{-3}\,du$.

   Let $u=\sec x+\tan x$. Then $du=(\sec x\tan x+\sec^2 x)\,dx=\sec x(\tan x+\sec x)\,dx=\sec x\,u\,dx$, so $\sec x\,dx=\dfrac{du}{u}$. Hence

   $\int\frac{\sec x}{(\sec x+\tan x)^2}\,dx=\int\frac{1}{u^2}\cdot\frac{du}{u}=\int u^{-3}\,du=-\frac{1}{2u^2}+C=\frac{-1}{2(\sec x+\tan x)^2}+C.$

   ✓Final answer

   The correct option is (B).

10. KEAM 2025Set eng-2025-04284 marksMCQ
    Q.$\int x(1 - x)^{10}\,dx =$ (A) $\frac{(1-x)^{12}}{12} - \frac{(1-x)^{11}}{11} + C$ (B) $\frac{(1-x)^{12}}{12} + \frac{(1-x)^{11}}{11} + C$ (C) $\frac{(1-x)^{11}}{11} - \frac{(1-x)^{10}}{10} + C$ (D) $\frac{(1-x)^{12}}{12} + \frac{(1+x)^{11}}{11} + C$ (E) $\frac{(1-x)^{12}}{12} - \frac{(1+x)^{11}}{11} + C$

    ›Reveal solutionSolution

    With $u=1-x$ the integral becomes $\dfrac{u^{12}}{12}-\dfrac{u^{11}}{11}$, i.e. $\dfrac{(1-x)^{12}}{12}-\dfrac{(1-x)^{11}}{11}+C$.

    Let $u=1-x$, $x=1-u$, $dx=-du$:

    $\int x(1-x)^{10}\,dx=-\int(1-u)u^{10}\,du=-\int(u^{10}-u^{11})\,du=\frac{u^{12}}{12}-\frac{u^{11}}{11}+C.$

    Restoring $u=1-x$:

    $=\frac{(1-x)^{12}}{12}-\frac{(1-x)^{11}}{11}+C.$

    ✓Final answer

    The correct option is (A).

11. 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).

12. KEAM 2026Set eng-2026-04184 marksMCQ
    Q.$\displaystyle\int \dfrac{10e^x}{(2e^x + 5)^3}\, dx$ is equal to (A) $\dfrac{5}{2(2e^x + 5)^2} + C$ (B) $\dfrac{-5}{(2e^x + 5)^2} + C$ (C) $\dfrac{-10}{(2e^x + 5)^2} + C$ (D) $\dfrac{-5}{2(2e^x + 5)^2} + C$ (E) $\dfrac{5}{(2e^x + 5)^2} + C$

    ›Reveal solutionSolution

    Substitute $u=2e^x+5$, $du=2e^x\,dx$, giving $5\int u^{-3}du=-\dfrac{5}{2(2e^x+5)^2}+C$.

    Let $u=2e^x+5$, so $du=2e^x\,dx$, i.e. $e^x\,dx=\tfrac{du}{2}$. Then

    $\int\dfrac{10e^x}{(2e^x+5)^3}dx=\int\dfrac{10}{u^3}\cdot\dfrac{du}{2}=5\int u^{-3}\,du=5\cdot\dfrac{u^{-2}}{-2}=-\dfrac{5}{2u^2}+C.$

    Hence the answer is $-\dfrac{5}{2(2e^x+5)^2}+C$.

    ✓Final answer

    The correct option is (D).