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

Showing the 12 most recent of 16 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.

   [!FORMULA] $\int \frac{d x}{x \sqrt{x^2+4}}=$

   (A) $\frac{1}{4} \log \left|\frac{\sqrt{x^2+4}-2}{\sqrt{x^2+4}+2}\right|+C$ (B) $\frac{1}{4} \log \left|\frac{\sqrt{x^2+4}+2}{\sqrt{x^2+4}-2}\right|+C$ (C) $\frac{1}{2} \log \left|\frac{\sqrt{x^2+4}+2}{\sqrt{x^2+4}-2}\right|+C$ (D) $\frac{1}{2} \log \left|\frac{\sqrt{x^2+4}-2}{\sqrt{x^2+4}+2}\right|+C$

   ›Reveal solutionSolution

   Substituting $x=2\tan\theta$ gives $\frac{1}{4}\log\left|\frac{\sqrt{x^2+4}-2}{\sqrt{x^2+4}+2}\right|+C$ — option (A).

   For $\displaystyle\int\frac{dx}{x\sqrt{x^2+4}}$ put $x=2\tan\theta$, so $dx=2\sec^2\theta\,d\theta$ and $\sqrt{x^2+4}=2\sec\theta$:

   $\int\frac{2\sec^2\theta\,d\theta}{2\tan\theta\cdot 2\sec\theta}=\frac{1}{2}\int\csc\theta\,d\theta=\frac{1}{2}\log\left|\csc\theta-\cot\theta\right|+C$

   With $\csc\theta=\dfrac{\sqrt{x^2+4}}{x}$ and $\cot\theta=\dfrac{2}{x}$:

   $=\frac{1}{2}\log\left|\frac{\sqrt{x^2+4}-2}{x}\right|+C$ …

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2. COMEDK 2021Set 20211 markMCQ
   Q.Integral of $\int {{{dx} \over {{x^2}{{[1 + {x^4}]}^{3/4}}}}}$. (A) $- 4{({x^{1/4}} + 1)^{1/4}} + C$ (B) $4{({x^{1/4}} + 1)^{1/4}} + C$ (C) $4{({x^4} + 1)^{1/4}} + C$ (D) None of these

   ›Reveal solutionSolution

   The result -(x^4 + 1)^(1/4)/x + C matches none of options (A), (B), (C) (they are missing the 1/x factor, and (A)/(B) even have x^(1/4)).

   Concept: for integrands of the form 1/(x^2 (1 + x^4)^(3/4)), take x^4 out of the bracket and substitute u = 1 + x^(-4).

   (1 + x^4)^(3/4) = x^3 (1 + x^(-4))^(3/4) (for x > 0).

   So the integrand = 1 / [ x^2 * x^3 * (1 + x^(-4))^(3/4) ] = x^(-5) (1 + x^(-4))^(-3/4).

   Let u = 1 + x^(-4) => du = -4 x^(-5) dx => x^(-5) dx = -du/4.

   I = -(1/4) * integral u^(-3/4) du = -(1/4) * (u^(1/4)/(1/4)) + C = -u^(1/4) + C

   = -(1 + x^(-4))^(1/4) + C

   = -((x^4 + 1)/x^4)^(1/4) + C

   = -(x^4 + 1)^(1/4) / x + C.

   Check by differentiating: d/dx [ -(1 + x^4)^(1/4) x^(-1) ] = -(1/4)(1 + x^4)^(-3/4)(4x^3)x^(-1) + (1 + x^4)^(1/4) x^(-2) …

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3. COMEDK 2023Set 2023-M1 markMCQ
   Q.$\int \frac{4^x}{\sqrt{1-16^x}} d x$ is equal to (A) $(\log 4) \sin ^{-1} 4^x+C$ (B) $\frac{1}{4} \sin ^{-1}\left(4^x\right)+C$ (C) $\frac{1}{\log 4} \sin ^{-1} 4^x+C$ (D) $4 \log 4 \sin ^{-1} 4+C$

   ›Reveal solutionSolution

   Put $u=4^x$ so $16^x=u^2$ and $du=4^x\ln4\,dx$; the integral becomes $\dfrac{1}{\ln4}\int\dfrac{du}{\sqrt{1-u^2}}=\dfrac{1}{\log4}\sin^{-1}(4^x)+C$.

   $\displaystyle\int\frac{4^x}{\sqrt{1-16^x}}\,dx.$ Let $u=4^x\Rightarrow du=4^x\ln4\,dx\Rightarrow 4^x\,dx=\dfrac{du}{\ln4}$, and $16^x=(4^x)^2=u^2$: …

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4. COMEDK 2024Set 2024-M1 markMCQ
   Q.

   [!FORMULA] $\text { The value of } \int \frac{1}{x+\sqrt{x-1}} d x \text { is }$

   (A) $\log (x+\sqrt{x-1})+\sin ^{-1}\left(\sqrt{\frac{x-1}{x}}\right)+C$ (B) $\log (x+\sqrt{x-1})-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \sqrt{x-1}+1}{\sqrt{3}}\right)+C$ (C) $\log (x+\sqrt{x-1})+C$ (D) $\log (x-1+\sqrt{x-1})+\frac{1}{\sqrt{3}} \log \left|\frac{x-2-\sqrt{3}}{x-2+\sqrt{3}}\right|+C$

   ›Reveal solutionSolution

   The integral simplifies by substituting $t = \sqrt{x-1}$, turning it into a rational function that integrates to a logarithm and an arctangent, matching option (B).

   We are asked to evaluate

   $\int \frac{1}{x + \sqrt{x-1}} \, dx.$

   The presence of $\sqrt{x-1}$ suggests a substitution that removes the square root, turning the integrand into a rational function. The trick is to set $t = \sqrt{x-1}$, so that $x = t^2 + 1$ and $dx = 2t \, dt$. This transforms the integral into a form we can handle with partial fractions or a standard arctangent formula.

   Let’s work through it step by step.

   1. Substitute $t = \sqrt{x-1}$. Then $x = t^2 + 1$ and $dx = 2t \, dt$. The denominator becomes

   $x + \sqrt{x-1} = (t^2 + 1) + t = t^2 + t + 1.$

   So the integral becomes

   $\int \frac{1}{t^2 + t + 1} \cdot 2t \, dt = 2 \int \frac{t}{t^2 + t + 1} \, dt.$

   2. Prepare for integration by rewriting the numerator to match the derivative of the denominator. The derivative of $t^2 + t + 1$ is $2t + 1$. We have $2t$ in the numerator, so write

   $2t = (2t + 1) - 1.$

   Then

   $2 \int \frac{t}{t^2 + t + 1} \, dt = \int \frac{2t + 1}{t^2 + t + 1} \, dt - \int \frac{1}{t^2 + t + 1} \, dt.$

   3. First integral:

   $\int \frac{2t + 1}{t^2 + t + 1} \, dt = \log|t^2 + t + 1| + C_1.$

   Since $t^2 + t + 1 > 0$ for all real $t$, we can drop the absolute value.

   4. Second integral: Complete the square in the denominator:

   $t^2 + t + 1 = \left(t + \frac12\right)^2 + \frac34.$

   So

   $\int \frac{1}{t^2 + t + 1} \, dt = \int \frac{1}{\left(t + \frac12\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \, dt.$

   Using the formula $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right)$, with $u = t + \frac12$ and $a = \frac{\sqrt{3}}{2}$, we get

   $\int \frac{1}{t^2 + t + 1} \, dt = \frac{2}{\sqrt{3}} \tan^{-1}\left( \frac{2t + 1}{\sqrt{3}} \right) + C_2.$

   5. Combine results: $$$$

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5. COMEDK 2021Set 20211 markMCQ
   Q.$\int {{{{2^x}} \over {\sqrt {1 - {4^x}} }}dx}$ is equal to (A) $(\log 2){\sin ^{ - 1}}{2^x} + C$ (B) ${1 \over 2}{\sin ^{ - 1}}{2^x} + C$ (C) ${1 \over {\log 2}}{\sin ^{ - 1}}{2^x} + C$ (D) $2\log 2{\sin ^{ - 1}}{2^x} + C$

   ›Reveal solutionSolution

   I = (1/log 2) * integral du / sqrt(1 - u^2) = (1/log 2) * arcsin(u) + C = (1/log 2) * arcsin(2^x) + C.

   Concept: substitution reducing the integrand to the standard form 1/sqrt(1 - u^2), whose integral is arcsin(u).

   I = integral 2^x / sqrt(1 - 4^x) dx. Note 4^x = (2^x)^2.

   Put u = 2^x. Then du = 2^x * log 2 dx, so 2^x dx = du / log 2. …

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6. COMEDK 2023Set 2023-M1 markMCQ
   Q.$\int \frac{x d x}{2(1+x)^{3 / 2}}$ is equal to (A) $\frac{2+x}{\sqrt{1+x}}+C$ (B) $\frac{2+x}{x \sqrt{1+x}}+C$ (C) $\frac{x}{\sqrt{1+x}}+C$ (D) $-\frac{x}{\sqrt{1+x}}+C$

   ›Reveal solutionSolution

   With $u=1+x$, the integral becomes $\tfrac12\int(u^{-1/2}-u^{-3/2})\,du=u^{1/2}+u^{-1/2}=\dfrac{2+x}{\sqrt{1+x}}+C$.

   $\displaystyle\int\frac{x\,dx}{2(1+x)^{3/2}}.$ Let $u=1+x\Rightarrow x=u-1,\ dx=du$:

   $\frac12\int\frac{u-1}{u^{3/2}}\,du=\frac12\int\left(u^{-1/2}-u^{-3/2}\right)du.$ …

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7. COMEDK 2026Set 2026-M1 markMCQ
   Q.

   [!FORMULA] $\int \frac{e^{\log \left(1+\frac{1}{x^2}\right)}}{x^2+\frac{1}{x^2}} d x=$

   (A) $\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^2+1}{x \sqrt{2}}\right)+C$ (B) $\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{2} x}\right)+C$ (C) $-\frac{1}{\sqrt{2}} \tan ^{-1}\left(x-\frac{1}{x}\right)+C$ (D) $\frac{1}{\sqrt{2}} \tan ^{-1}\left(x-\frac{1}{x}\right)+C$

   ›Reveal solutionSolution

   The integrand simplifies dramatically using exponent rules and algebraic manipulation, leading to a standard arctangent integral; the correct antiderivative matches option (D).

   We start with the integral

   $\int \frac{e^{\log \left(1+\frac{1}{x^2}\right)}}{x^2+\frac{1}{x^2}} \, dx.$

   Concept & Intuition

   The presence of $e^{\log(\cdots)}$ is a huge clue: for any positive argument, $e^{\log(u)} = u$. That immediately collapses the numerator into something algebraic. Then the denominator is symmetric in $x$ and $1/x$, which often suggests a substitution like $t = x - 1/x$ because its derivative appears in the numerator. This is a classic trick for integrals involving $x^2 + 1/x^2$.

   ---

   Step-by-step solution

   1. Simplify the exponential Since $e^{\log(u)} = u$ for $u>0$ (and $1+1/x^2 > 0$ for all real $x \neq 0$), we have

   $e^{\log\left(1+\frac{1}{x^2}\right)} = 1 + \frac{1}{x^2}.$

   So the integral becomes

   $\int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} \, dx.$

   2. Rewrite numerator and denominator Multiply numerator and denominator by $x^2$ to clear fractions:

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

   So the integral is

   $\int \frac{x^2 + 1}{x^4 + 1} \, dx.$

   3. Divide numerator and denominator by $x^2$ This is the key algebraic trick:

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

   Notice that the numerator $1 + 1/x^2$ is the derivative of $x - 1/x$ (since $\frac{d}{dx}(x - 1/x) = 1 + 1/x^2$).

   Also, $x^2 + 1/x^2 = (x - 1/x)^2 + 2$.

   4. Substitute Let $t = x - \frac{1}{x}$. Then

   $dt = \left(1 + \frac{1}{x^2}\right) dx.$

   And

   $x^2 + \frac{1}{x^2} = t^2 + 2.$

   The integral becomes

   $\int \frac{dt}{t^2 + 2}.$

   5. Integrate This is a standard arctangent form:

   $\int \frac{dt}{t^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{t}{a}\right) + C.$

   Here $a = \sqrt{2}$, so

   $\int \frac{dt}{t^2 + 2} = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t}{\sqrt{2}}\right) + C.$

   6. Back-substitute Replace $t$ with $x - \frac{1}{x}$:

   $\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x - \frac{1}{x}}{\sqrt{2}}\right) + C.$

   Simplify the argument:

   $\frac{x - \frac{1}{x}}{\sqrt{2}} = \frac{x^2 - 1}{\sqrt{2}\, x}.$

   So the antiderivative is

   $\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x^2 - 1}{\sqrt{2}\, x}\right) + C.$ …

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8. COMEDK 2024Set 2024-A1 markMCQ
   Q.

   [!FORMULA] $\int \frac{f^{\prime}(x)}{f(x) \log (f(x))} d x \text { is equal to }$

   (A) $f(x) \log f(x)+C$ (B) $\frac{1}{\log (\log f(x))}+C$ (C) $\frac{f(x)}{\log f(x)}+C$ (D) $\log (\log f(x))+C$

   ›Reveal solutionSolution

   The integral simplifies by substitution $u = \log(f(x))$, leading directly to $\log(\log f(x)) + C$, so the correct choice is (D).

   Concept & Intuition

   When you see a fraction where the numerator is the derivative of the denominator’s “inside,” a substitution is almost always the cleanest path. Here, the denominator is $f(x) \log(f(x))$, and the numerator is $f'(x)$. Notice that the derivative of $\log(f(x))$ is $\frac{f'(x)}{f(x)}$, which appears in the integrand. That suggests setting $u = \log(f(x))$, turning the whole integral into a simple $\int \frac{du}{u}$.

   Step-by-step solution

   1. Identify the substitution Let $u = \log(f(x))$. Then differentiate:

   $\frac{du}{dx} = \frac{f'(x)}{f(x)} \quad\Rightarrow\quad du = \frac{f'(x)}{f(x)}\,dx.$

   2. Rewrite the integral The original integral is

   $\int \frac{f'(x)}{f(x) \log(f(x))}\,dx.$

   Factor the $f(x)$ in the denominator:

   $\int \frac{f'(x)/f(x)}{\log(f(x))}\,dx.$

   Now substitute $u = \log(f(x))$ and $du = \frac{f'(x)}{f(x)}dx$:

   $\int \frac{du}{u}.$

   3. Integrate The integral $\int \frac{du}{u}$ is a standard result:

   $\int \frac{du}{u} = \log|u| + C.$

   4. Back-substitute Replace $u$ with $\log(f(x))$: $\log|\log(f(x))| + C.$ …

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9. COMEDK 2021Set 2021-B1 markMCQ
   Q.$\int \sqrt{\frac{\cos x - \cos^3 x}{1 - \cos^3 x}}\, dx =$ (A) $-\frac{1}{3}\log\left|\frac{1 + \cos^{3/2} x}{1 - \cos^{3/2} x}\right| + c$ (B) $-\frac{1}{3}\log\left|\frac{\cos^{3/2} x - 1}{\cos^{3/2} x + 1}\right| + c$ (C) $-\frac{2}{3}\sin^{-1}(\cos^{3/2} x) + c$ (D) $-\frac{2}{3}\sin^{-1}(\cos^3 x) + c$

   ›Reveal solutionSolution

   The integral equals $-\tfrac23\sin^{-1}(\cos^{3/2}x)+c$.

   Simplify the radicand: $\cos x-\cos^3x=\cos x(1-\cos^2x)=\cos x\sin^2x$, so

   $\sqrt{\frac{\cos x-\cos^3x}{1-\cos^3x}}=\frac{\sqrt{\cos x}\,|\sin x|}{\sqrt{1-\cos^3x}}.$

   Let $u=\cos^{3/2}x$. Then $\dfrac{du}{dx}=\tfrac32\cos^{1/2}x\cdot(-\sin x)=-\tfrac32\sqrt{\cos x}\,\sin x$, so $\sqrt{\cos x}\,\sin x\,dx=-\tfrac23\,du$, and $1-\cos^3x=1-u^2$.

   Thus …

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10. COMEDK 2022Set 20221 markMCQ
    Q.$\int \frac{3^x}{\sqrt{1-9^x}} d x$ is equal to (A) $(\log 3) \sin ^{-1} 3^x+C$ (B) $\frac{1}{3} \sin ^{-1}\left(3^x\right)+C$ (C) $\frac{1}{\log 3} \sin ^{-1} 3^x+C$ (D) $3 \log 3 \sin ^{-1} 3^x+C$

    ›Reveal solutionSolution

    I = (1/log 3) * Integral du / sqrt(1 - u^2) = (1/log 3) * arcsin(u) + C = (1 / log 3) * sin^-1 (3^x) + C

    Concept: Substitution reducing to the arcsin form, integral du/sqrt(1 - u^2) = arcsin u.

    I = Integral of 3^x / sqrt(1 - 9^x) dx , and 9^x = (3^x)^2

    Put u = 3^x => du = 3^x (log 3) dx => 3^x dx = du / log 3

    I = (1/log 3) * Integral du / sqrt(1 - u^2) …

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11. COMEDK 2025Set 2025-A1 markMCQ
    Q.$\int \frac{e^{\tan ^{-1} x}}{\left(1+x^2\right)}\left(1+x+x^2\right) d x=$ (A) $e^{\tan ^{-1} x}+c$ (B) $x e^{\tan ^{-1} x}+c$ (C) $\frac{e^{\tan ^{-1} x}}{\left(1+x^2\right)}+c$ (D) $\frac{x e^{\tan ^{-1} x}}{\left(1+x^2\right)}+c$

    ›Reveal solutionSolution

    The integral simplifies by substituting $u = \tan^{-1} x$, which turns the expression into a sum of a standard exponential integral and a derivative-of-product pattern, yielding $x e^{\tan^{-1} x} + C$. The correct option is (B).

    ---

    The key insight is that the denominator $1+x^2$ is exactly the derivative of $\tan^{-1} x$, so the substitution $u = \tan^{-1} x$ is natural. Once we do that, the polynomial $1+x+x^2$ becomes something in terms of $\tan u$, and we can split the integral into two recognizable pieces.

    1. Substitute $u = \tan^{-1} x$. Then $du = \frac{dx}{1+x^2}$, and $x = \tan u$. The integral becomes

    $\int e^{\tan^{-1} x} \cdot \frac{1+x+x^2}{1+x^2}\,dx = \int e^{u} \left(1 + \tan u + \tan^2 u\right) du.$

    2. Simplify the trigonometric expression. Recall $1 + \tan^2 u = \sec^2 u$. So

    $1 + \tan u + \tan^2 u = \sec^2 u + \tan u.$

    The integral is now

    $\int e^{u} (\sec^2 u + \tan u)\, du.$

    3. Split and recognize patterns.

    $\int e^{u} \sec^2 u \, du + \int e^{u} \tan u \, du.$

    Notice that $\frac{d}{du}(\tan u) = \sec^2 u$. The first integral is of the form $\int e^{u} f'(u) du$ with $f(u) = \tan u$, and the second is $\int e^{u} f(u) du$.

    4. Use the product rule in reverse. For any differentiable $f(u)$,

    $\frac{d}{du}\left( e^{u} f(u) \right) = e^{u} f'(u) + e^{u} f(u).$

    Here $f(u) = \tan u$, so

    $$$$

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12. COMEDK 2021Set 2021-B1 markMCQ
    Q.If $\displaystyle\int\dfrac{2^x}{\sqrt{1-4^x}}\,dx=k\sin^{-1}(2^x)+c$, then k is (A) $\dfrac{\log 2}{2}$ (B) $\dfrac{1}{\log 2}$ (C) $\dfrac{1}{2\log 2}$ (D) $\log 2$

    ›Reveal solutionSolution

    $k=\dfrac{1}{\log 2}$.

    Let $u=2^x$, so $du=2^x\ln2\,dx$ and $4^x=u^2$. Then

    $\int\frac{2^x}{\sqrt{1-4^x}}\,dx=\int\frac{u}{\sqrt{1-u^2}}\cdot\frac{du}{u\ln2}=\frac{1}{\ln2}\int\frac{du}{\sqrt{1-u^2}}=\frac{1}{\ln2}\sin^{-1}u+c.$ …

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