Q.∫ 𝒅𝒙 𝒙𝟑(𝟏+𝒙𝟒) 𝟏 𝟐 equals
(A) − 1 2𝑥2 √1 + 𝑥4 + 𝑐
(B) 1 2𝑥 √1 + 𝑥4 + 𝑐
(C) − 1 4𝑥 √1 + 𝑥4 + 𝑐
(D) 1 4𝑥2 √1 + 𝑥4 + 𝑐

Showing the 12 most recent of 32 on this concept.

1. 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

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

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2. KEAM 2022Set eng-2022-P2-B14 marksMCQ
   Q.$\int \frac{1}{x^3}\sqrt{1 - \frac{1}{x^2}} \, dx =$ (A) $\frac{-1}{6}\left(1 - \frac{1}{x^2}\right)^{\frac{3}{2}} + C$ (B) $\frac{1}{3}\left(1 - \frac{1}{x^2}\right)^{\frac{3}{2}} + C$ (C) $\frac{-1}{3}\left(1 - \frac{1}{x^2}\right)^{\frac{3}{2}} + C$ (D) $\frac{4}{3}\left(1 - \frac{1}{x^2}\right)^{\frac{3}{2}} + C$ (E) $\frac{-4}{3}\left(1 - \frac{1}{x^2}\right)^{\frac{3}{2}} + C$

   ›Reveal solutionSolution

   The integral equals $\frac13\left(1-\frac1{x^2}\right)^{3/2}+C$.

   Concept and Intuition

   The derivative of $1-\frac1{x^2}$ is $\frac2{x^3}$, which matches the $\frac1{x^3}$ factor outside the root, so a substitution linearizes the integral.

   Step-by-Step Solution

   1. Let $u=1-\frac1{x^2}$.
   2. $du=\frac{2}{x^3}\,dx\Rightarrow \frac{dx}{x^3}=\frac{du}{2}$.
   3. $\int \frac{1}{x^3}\sqrt{1-\frac1{x^2}}\,dx=\int \sqrt u\,\frac{du}{2}=\frac12\cdot\frac{u^{3/2}}{3/2}$.
   4. $=\frac12\cdot\frac23 u^{3/2}=\frac13 u^{3/2}=\frac13\left(1-\frac1{x^2}\right)^{3/2}+C$. …

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3. 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 …

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4. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.$\int x^7 (x^8 + 1)^{-3/4}\, dx =$ (A) $\dfrac{1}{2}\left(1 + \dfrac{1}{x^8}\right)^{1/4} + C$ (B) $4\left(1 + \dfrac{1}{x^8}\right)^{1/4} + C$ (C) $(x^8 + 1)^{1/4} + C$ (D) $4(x^8 + 1)^{1/4} + C$ (E) $\dfrac{1}{2}(x^8 + 1)^{1/4} + C$

   ›Reveal solutionSolution

   Substituting u=x^8+1 gives (1/2)(x^8+1)^{1/4} + C.

   Concept and Intuition

   The presence of x^7 alongside x^8 signals the substitution u = x^8 + 1, whose differential absorbs x^7 dx.

   Step-by-Step Solution

   1. Let u = x^8 + 1, then du = 8 x^7 dx, so x^7 dx = du/8.
   2. Integral = (1/8) integral of u^{-3/4} du.
   3. = (1/8) * u^{1/4}/(1/4) = (1/8)4u^{1/4} = (1/2)u^{1/4}. …

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5. KEAM 2025Set eng-2025-04284 marksMCQ
   Q.$\int \frac{dx}{\cos^{2/3}x \, \sin^{4/3}x}$ is (A) $3\tan^3 x + C$ (B) $3\tan^{1/3} x + C$ (C) $-3\tan^{1/3} x + C$ (D) $-3\tan^{-1/3} x + C$ (E) $3\tan^{-1/3} x + C$

   ›Reveal solutionSolution

   Rewrite as $\sec^2x\,\tan^{-4/3}x\,dx$; sub $t=\tan x$ to get $\int t^{-4/3}dt=-3t^{-1/3}=-3\tan^{-1/3}x+C$.

   The integrand $\dfrac{1}{\cos^{2/3}x\,\sin^{4/3}x}$ has denominator powers summing to $2$, so factor out $\cos^2x$. Multiplying numerator and denominator by $\cos^{4/3}x$ (equivalently dividing top and bottom by $\cos^2x$): …

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

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7. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.$\displaystyle\int \left(27x^3\left(1 - x^3\right)\right)^{\frac{2}{3}}\, dx =$ (A) $-\dfrac{3}{4}\left(1 - x^3\right)^{\frac{4}{3}} + C$ (B) $-\dfrac{3}{5}\left(1 - x^3\right)^{\frac{5}{3}} + C$ (C) $-\dfrac{9}{3}\left(1 - x^3\right)^{\frac{1}{3}} + C$ (D) $-\dfrac{9}{4}\left(1 - x^3\right)^{\frac{4}{3}} + C$ (E) $-\dfrac{9}{5}\left(1 - x^3\right)^{\frac{5}{3}} + C$

   ›Reveal solutionSolution

   Simplify to $9x^2(1-x^3)^{2/3}$, then substitute $u=1-x^3$ to obtain $-\dfrac{9}{5}(1-x^3)^{5/3}+C$.

   Since $\bigl(27x^3(1-x^3)\bigr)^{2/3}=27^{2/3}(x^3)^{2/3}(1-x^3)^{2/3}=9x^2(1-x^3)^{2/3}$, let $u=1-x^3$, $du=-3x^2\,dx$, so $x^2\,dx=-\tfrac{du}{3}$: …

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8. 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 $du=\frac{4x}{\sqrt{4x^2+7}}\,dx$. …

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9. 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 …

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10. 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 …

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11. KEAM 2024Set eng-2024-06074 marksMCQ
    Q.$\displaystyle\int x^5 e^{x^3}\,dx =$ (A) $\dfrac{e^{x^3}}{3}(x^3-1)+C$ (B) $\dfrac{e^{x^3}}{5}(x^5-1)+C$ (C) $\dfrac{e^{x^3}}{4}(x^4-1)+C$ (D) $\dfrac{e^{x^3}}{3}(x^5-1)+C$ (E) $\dfrac{x^3 e^{x^3}}{3}+C$

    ›Reveal solutionSolution

    Substitute $u=x^3$, then integrate $ue^u$ by parts.

    Let $u=x^3$, so $du=3x^2dx$ and $x^5dx=x^3\cdot x^2dx=\frac{u}{3}du$. Thus

    $\int x^5e^{x^3}dx=\frac13\int u\,e^u\,du.$

    By parts, $\int ue^u du=ue^u-e^u=e^u(u-1)$, so …

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12. KEAM 2024Set eng-2024-06054 marksMCQ
    Q.$\int \frac{dx}{x^8 \left(\frac{1}{x^7} + 1\right)^{2/3}}$ is equal to (A) $\frac{3}{7}\left(\frac{1}{x^7} + 1\right)^{2/3} + C$ (B) $-\frac{3}{7}\left(\frac{1}{x^7} + 1\right)^{2/3} + C$ (C) $-\frac{3}{7}\left(\frac{1}{x^7} + 1\right)^{1/3} + C$ (D) $\frac{3}{7}\left(\frac{1}{x^7} + 1\right)^{1/3} + C$ (E) $\frac{7}{3}\left(\frac{1}{x^7} + 1\right)^{2/3} + C$

    ›Reveal solutionSolution

    The substitution $u=x^{-7}+1$ reduces it to $-\frac17\int u^{-2/3}\,du$.

    Let $u=\dfrac{1}{x^7}+1$. Then $du=-\dfrac{7}{x^8}\,dx$, so $\dfrac{dx}{x^8}=-\dfrac{du}{7}$. The integral becomes …

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