Q.Find $\int \dfrac{(x^4 - x)^{1/4}}{x^5}\, dx$

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

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

   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).

2. 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

   $=\frac13 e^{x^3}(x^3-1)+C=\frac{e^{x^3}}{3}(x^3-1)+C.$

   ✓Final answer

   The correct option is (A).

3. 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}.
   4. = (1/2)(x^8+1)^{1/4} + C.

   Common Mistakes

   • Dropping the 1/8 from du and getting 4(x^8+1)^{1/4} instead.

   ✓Final answer

   The correct option is (E) — (1/2)(x^8+1)^{1/4} + C.

   ANSWER: E

4. 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 $x^2\,dx=-\tfrac{du}{3}$:

   $\int 9x^2(1-x^3)^{2/3}dx=9\left(-\tfrac13\right)\int u^{2/3}du=-3\cdot\dfrac{u^{5/3}}{5/3}=-\dfrac{9}{5}u^{5/3}+C.$

   Thus the integral is $-\dfrac{9}{5}(1-x^3)^{5/3}+C$.

   ✓Final answer

   The correct option is (E).

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. $=\frac12\cdot\frac23 u^{3/2}=\frac13 u^{3/2}=\frac13\left(1-\frac1{x^2}\right)^{3/2}+C$.

   Common Mistakes

   • Getting the sign or factor of $du$ wrong (it is $+\frac2{x^3}$).
   • Forgetting to halve after substituting.

   ✓Final answer

   The correct option is (B) — $\frac{1}{3}\left(1 - \frac{1}{x^2}\right)^{\frac{3}{2}} + C$.

   ANSWER: B

6. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.$\int\dfrac{\sec^2\left(\sqrt{2x+5}\right)}{\sqrt{2x+5}}\,dx=$ (A) $2\tan\left(\sqrt{2x+5}\right)+C$ (B) $\dfrac{1}{2}\tan\left(\sqrt{2x+5}\right)+C$ (C) $\tan\left(2x+5\right)+C$ (D) $\tan\left(\sqrt{2x+5}\right)+C$ (E) $2\tan\left(2x+5\right)+C$

   ›Reveal solutionSolution

   Substitute $u=\sqrt{2x+5}$; the differential cancels the $\dfrac{1}{\sqrt{2x+5}}$ factor.

   Let $u=\sqrt{2x+5}$. Then

   $du=\dfrac{1}{2\sqrt{2x+5}}\cdot 2\,dx=\dfrac{dx}{\sqrt{2x+5}}.$

   So the integral becomes

   $\int \sec^{2}u\,du=\tan u+C=\tan\left(\sqrt{2x+5}\right)+C.$

   ✓Final answer

   The correct option is (D).

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

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

   ✓Final answer

   The correct option is (A).

8. KEAM 2024Set eng-2024-06064 marksMCQ
   Q.$\int_{0}^{\pi/4}\left(\tan^{3} x+\tan^{5} x\right)dx =$ (A) $\frac{5}{12}$ (B) $\frac{1}{3}$ (C) $\frac{1}{4}$ (D) $\frac{1}{6}$ (E) $\frac{1}{12}$

   ›Reveal solutionSolution

   Factor out $\tan^3x\sec^2x$ and substitute $u=\tan x$ to get $\frac{1}{4}$.

   $\tan^3x+\tan^5x=\tan^3x(1+\tan^2x)=\tan^3x\sec^2x$.

   Let $u=\tan x$, $du=\sec^2x\,dx$. Limits: $x=0\to u=0$, $x=\frac{\pi}{4}\to u=1$.

   $\int_0^{\pi/4}\tan^3x\sec^2x\,dx=\int_0^1u^3\,du=\left[\frac{u^4}{4}\right]_0^1=\frac{1}{4}$.

   ✓Final answer

   The correct option is (C).

9. 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

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 $=\displaystyle\int e^{u}\left(-\tfrac16\right)du=-\tfrac16 e^{u}+C$.
    3. Back-substitute: $=-\dfrac16 e^{1-x^6}+C$.

    Common Mistakes

    • Dropping the negative sign from $du=-6x^5\,dx$.

    ✓Final answer

    The correct option is (C) — $\dfrac{-1}{6} e^{1-x^6} + C$.

    ANSWER: C

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 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).