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 44 on this concept.

1. CBSE 2026Set 65/1/11 markMCQ
   Q.If $\int \frac{3ax}{b^2 + c^2x^2} dx = A \log |b^2 + c^2x^2| + K$, then the value of A is: (A) $3a$ (B) $\frac{3a}{2b^2}$ (C) $\frac{3a}{b^2c^2}$ (D) $\frac{3a}{2c^2}$

   ›Reveal solutionSolution

   The integral fits the pattern $\int \frac{du}{u} = \log|u| + C$ after a substitution. The constant $A$ turns out to be $\frac{3a}{2c^2}$, which corresponds to option (D).

   The problem gives you the result of an integral and asks you to identify the constant $A$ that makes the equation true. This is a classic "match the form" question — you don't need to guess; you just need to perform the integration carefully and compare.

   The key insight is that the integrand $\frac{3ax}{b^2 + c^2x^2}$ is a rational function where the numerator is almost the derivative of the denominator. The derivative of $b^2 + c^2x^2$ is $2c^2x$. Our numerator is $3ax$, which is a constant multiple of $x$. So a simple substitution $u = b^2 + c^2x^2$ will turn the integral into $\int \frac{du}{u}$.

   Let's work through it step by step.

   1. Set up the substitution.

      Let $u = b^2 + c^2x^2$. Then $du = 2c^2x \, dx$, so $x \, dx = \frac{du}{2c^2}$.

   2. Rewrite the integral in terms of $u$.

      The integral is $\int \frac{3ax}{b^2 + c^2x^2} \, dx = \int \frac{3a}{u} \cdot (x \, dx)$.

      Substitute $x \, dx = \frac{du}{2c^2}$:

   $\int \frac{3a}{u} \cdot \frac{du}{2c^2} = \frac{3a}{2c^2} \int \frac{du}{u}.$

   3. Integrate. $\int \frac{du}{u} = \log|u| + C$, so

   $\frac{3a}{2c^2} \log|u| + C = \frac{3a}{2c^2} \log|b^2 + c^2x^2| + K,$

   where $K$ is the constant of integration (we renamed $C$ to $K$ to match the problem).

   4. Compare with the given form. …

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2. CBSE 2020Set 65/1/11 mark
   Q.Evaluate: $\int x^4 \log x \, dx$

   (OR)

   Evaluate: $\int \frac{2x}{\sqrt[3]{x^2+1}} \, dx$

   ›Reveal solutionSolution

   1. $\displaystyle\int x^4\log x\,dx=\tfrac{x^5}{5}\log x-\tfrac{x^5}{25}+C$.
   2. $\displaystyle\int\frac{2x}{\sqrt[3]{x^2+1}}\,dx=\tfrac32(x^2+1)^{2/3}+C$.

   Part (a)

   Use integration by parts, $\displaystyle\int u\,dv=uv-\int v\,du$, choosing $u=\log x$ (differentiates simply) and $dv=x^4\,dx$, so $du=\tfrac1x\,dx$ and $v=\dfrac{x^5}{5}$:

   $\int x^4\log x\,dx=\frac{x^5}{5}\log x-\int\frac{x^5}{5}\cdot\frac{1}{x}\,dx=\frac{x^5}{5}\log x-\frac15\int x^4\,dx.$ …

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3. CBSE 2020Set 65/1/11 mark
   Q.Find : $\int \frac{\text{dx}}{\sqrt{9-4x^2}}$

   ›Reveal solutionSolution

   The integral $\int \frac{dx}{\sqrt{9-4x^2}}$ is a standard inverse sine form. By rewriting the denominator as $\sqrt{4\left(\frac{9}{4} - x^2\right)}$ and using substitution $u = 2x$, we get the result $\frac{1}{2} \sin^{-1}\left(\frac{2x}{3}\right) + C$.

   When you see a square root with a constant minus a square term, your mind should immediately jump to the inverse trigonometric integrals. The classic formula is:

   $\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}\left(\frac{u}{a}\right) + C$

   Our job is to force the given integral into this exact shape. The denominator is $\sqrt{9 - 4x^2}$. Notice that $9 = 3^2$, so we have $a = 3$ in the formula. But the $4x^2$ term is not a pure $u^2$ — it has a coefficient 4. That’s the only obstacle.

   The key insight: Factor out the 4 from inside the square root. Write:

   $\sqrt{9 - 4x^2} = \sqrt{4\left(\frac{9}{4} - x^2\right)} = 2 \sqrt{\frac{9}{4} - x^2}$

   Now the integral becomes:

   $\int \frac{dx}{2 \sqrt{\frac{9}{4} - x^2}} = \frac{1}{2} \int \frac{dx}{\sqrt{\left(\frac{3}{2}\right)^2 - x^2}}$

   This is exactly the inverse sine form with $a = \frac{3}{2}$ and $u = x$. So:

   $\frac{1}{2} \sin^{-1}\left(\frac{x}{3/2}\right) + C = \frac{1}{2} \sin^{-1}\left(\frac{2x}{3}\right) + C$

   That’s the answer. But let’s walk through it step by step with a substitution to make it foolproof.

   1. Identify the target form. We want $\int \frac{du}{\sqrt{a^2 - u^2}}$. Here, the denominator has $\sqrt{9 - 4x^2}$. Compare with $a^2 - u^2$: we need $a^2 = 9$ and $u^2 = 4x^2$. So set $u = 2x$. Then $du = 2\,dx$, so $dx = \frac{du}{2}$.

   2. Substitute. The integral becomes:

      $\int \frac{dx}{\sqrt{9 - 4x^2}} = \int \frac{du/2}{\sqrt{9 - u^2}} = \frac{1}{2} \int \frac{du}{\sqrt{9 - u^2}}$ …

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4. CBSE 2026Set CX1 mark
   Q.Find the value of the integral $\int x^{2}\tan(x^{3}+2)\,dx$.

   ›Reveal solutionSolution

   Substitute $u=x^{3}+2$; the integral becomes $\tfrac13\int\tan u\,du=\tfrac13\ln|\sec(x^{3}+2)|+C$.

   Concept: The factor $x^{2}$ is (up to a constant) the derivative of the inner function $x^{3}+2$, so substitution works.

   Let $u=x^{3}+2\Rightarrow du=3x^{2}\,dx\Rightarrow x^{2}\,dx=\dfrac{du}{3}$.

   …

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5. CBSE 2026Set A1 markMCQ
   Q.$\int\frac{\tan(\sin^{-1}x)}{\sqrt{1-x^2}}\,dx =$

   (a) $\log|\sec(\sin^{-1}x)| + k$

   (b) $\log|\cos(\sin^{-1}x)| + k$

   (c) $\tan(\sin^{-1}x) + k$

   (d) $\log|\sin^{-1}x| + k$

   ›Reveal solutionSolution

   With $u=\sin^{-1}x$ (so $du=\tfrac{dx}{\sqrt{1-x^2}}$) the integral is $\int\tan u\,du=\log|\sec u|+k$.

   Let $u=\sin^{-1}x$. Then $du=\dfrac{dx}{\sqrt{1-x^2}}$, so

   …

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6. CBSE 2026Set A1 markMCQ
   Q.$\int\frac{dx}{e^x + e^{-x}} =$

   (a) $\cot^{-1}(e^x) + k$

   (b) $\tan^{-1}(e^x) + k$

   (c) $\log|e^x + 1| + k$

   (d) $\sin^{-1}(e^x) + k$

   ›Reveal solutionSolution

   Substitute $t=e^x$: $\int\tfrac{dx}{e^x+e^{-x}}=\int\tfrac{dt}{1+t^2}=\tan^{-1}(e^x)+k$.

   Multiply numerator and denominator by $e^x$:

   $\dfrac{1}{e^x+e^{-x}}=\dfrac{e^x}{e^{2x}+1}$.

   Let $t=e^x$, $dt=e^x\,dx$. Then

   …

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7. CBSE 2026Set ANNUAL1 markMCQ
   Q.$\int \sin(2x+3)\,dx =$

   (a) $\cos(2x+3)+C$

   (b) $-\dfrac{\cos(2x+3)}{2}+C$

   (c) $\tan 2x + C$

   (d) None of these

   ›Reveal solutionSolution

   $\int \sin(ax+b)\,dx = -\dfrac{\cos(ax+b)}{a}+C$.

   …

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8. CBSE 2026Set ANNUAL1 markMCQ
   Q.$\displaystyle\int_1^{e} \dfrac{(\log x)^2}{x}\, dx =$

   (a) $\dfrac{1}{3}e^3$

   (b) $\dfrac{1}{3}(e^3-1)$

   (c) $\dfrac{1}{3}$

   (d) None of these

   ›Reveal solutionSolution

   Substitute $u = \log x$, $du = dx/x$, converting the limits from $x=1,e$ to $u=0,1$.

   Let $u=\log x \Rightarrow du = \dfrac{dx}{x}$. When $x=1, u=0$; when $x=e, u=1$.

   …

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9. CBSE 2026Set ANNUAL1 markMCQ
   Q.$\displaystyle\int \dfrac{1}{x(1+\log x)}\,dx$ is equal to:

   (a) $x + \log x + c$

   (b) $|x + \log x| + c$

   (c) $\log|1 + \log x| + c$

   (d) $\log(1+x) + c$

   ›Reveal solutionSolution

   Substitute $u=1+\log x$ so $du=\dfrac{dx}{x}$, turning the integral into $\int\dfrac{du}{u}$.

   $I=\displaystyle\int\dfrac{1}{x(1+\log x)}\,dx$

   Let $u=1+\log x \Rightarrow du=\dfrac{1}{x}dx$.

   …

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10. CBSE 2026Set ANNUAL1 markMCQ
    Q.$\displaystyle\int \dfrac{\sin^6 x}{\cos^8 x}\,dx$ is equal to:

    ›Reveal solutionSolution

    Rewrite the integrand as $\tan^6x\sec^2x$ and substitute $t=\tan x$.

    $I=\displaystyle\int\dfrac{\sin^6x}{\cos^8x}\,dx=\int\dfrac{\sin^6x}{\cos^6x}\cdot\dfrac{1}{\cos^2x}\,dx=\int\tan^6x\sec^2x\,dx$

    Let $t=\tan x \Rightarrow dt=\sec^2x\,dx$. …

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11. CBSE 2026Set ANNUAL1 markMCQ
    Q.\int x^2 e^{x^3} dx equals:

    (a)

    (i) \frac{e^{x^3}}{3} + c

    (b)

    (ii) 3e^{x^3} + c

    (c)

    (iii) \frac{e^{x^2}}{3} + c

    (d)

    (iv) \frac{1}{2}e^{x^2} + c

    ›Reveal solutionSolution

    $\displaystyle\int x^2 e^{x^3}\,dx=\dfrac{e^{x^3}}{3}+c$ — option (i).

    Concept. Substitution ($u$-substitution): choose $u$ whose derivative already appears (up to a constant) in the integrand.

    Steps.

    • Let $u=x^3$, then $du=3x^2\,dx$, i.e. $x^2\,dx=\dfrac{1}{3}\,du$. …

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12. CBSE 2025Set ANNUAL1 markMCQ
    Q.$\displaystyle\int \dfrac{\cos x}{1+\sin^2 x}\, dx =$

    (a) $-\tan^{-1}(\sin x) + c$

    (b) $\tan^{-1}(\cos x) + c$

    (c) $\tan^{-1}(\sin x) + c$

    (d) $-\tan^{-1}(\cos x) + c$

    ›Reveal solutionSolution

    A direct substitution u = sin x reduces this to the standard ∫du/(1+u²) form.

    Let $u=\sin x$, so $du = \cos x\,dx$.

    …

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