Q.Integrate the following function: $\frac{6x+7}{\sqrt{(x-5)(x-4)}}$

Showing the 12 most recent of 32 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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7. 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

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

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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: $x=0\to u=0$, $x=\frac{\pi}{4}\to u=1$. …

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9. 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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10. 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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11. 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 …

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12. 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 $u=\sqrt t$, $t=u^2$, $dt=2u\,du$: …

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