Q.Integrate the following function: $(x^3-1)^{1/3} x^5$

Showing the 12 most recent of 32 on this concept.

1. 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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2. 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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3. 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 $x^2$:

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

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4. 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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5. 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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6. 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 …

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7. 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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8. KEAM 2025Set eng-2025-04264 marksMCQ
   Q.If $\displaystyle\int \dfrac{1}{x^7 \left(\dfrac{1}{x^6}+1\right)^{2/3}} \, dx = -\dfrac{1}{2}\left(\dfrac{1}{\dfrac{1}{x^6}+1}\right)^{p} + c$, then $p =$ (A) $\dfrac{2}{3}$ (B) $\dfrac{-1}{3}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{-2}{3}$ (E) $\dfrac{1}{6}$

   ›Reveal solutionSolution

   Substitute $u=\tfrac1{x^6}+1$; the result is $-\tfrac12u^{1/3}$, matched to the given form $-\tfrac12(1/u)^{p}$ gives $p=-1/3$.

   Let $u=\dfrac{1}{x^6}+1$, so $du=-\dfrac{6}{x^7}\,dx$, i.e. $\dfrac{dx}{x^7}=-\dfrac{du}{6}$.

   $\int \frac{u^{-2/3}}{x^7}\,dx=\int u^{-2/3}\left(-\frac{du}{6}\right)=-\frac16\cdot\frac{u^{1/3}}{1/3}=-\frac12u^{1/3}+c.$ …

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9. 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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10. KEAM 2021Set eng-2021-P2-B14 marksMCQ
    Q.$\int x^5 e^{1-x^6} \, dx =$ (A) $\dfrac{1}{6} e^{1-x^6} + C$ (B) $-e^{1-x^6} + C$ (C) $\dfrac{-1}{6} e^{1-x^6} + C$ (D) $\dfrac{x^5}{5} e^{1-x^6} + C$ (E) $\dfrac{x^6}{6} e^{1-x^6} + C$

    ›Reveal solutionSolution

    $\int x^5 e^{1-x^6}\,dx=-\dfrac16 e^{1-x^6}+C$.

    Concept and Intuition

    The exponent's derivative $\dfrac{d}{dx}(1-x^6)=-6x^5$ matches the algebraic factor $x^5$, so a $u$-substitution collapses the integral.

    Step-by-Step Solution

    1. Let $u=1-x^6$, then $du=-6x^5\,dx$, i.e. $x^5\,dx=-\dfrac16\,du$.
    2. Integral $=\displaystyle\int e^{u}\left(-\tfrac16\right)du=-\tfrac16 e^{u}+C$. …

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11. 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. …

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12. 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 $u=1-x$, $x=1-u$, $dx=-du$: …

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