Q.Evaluate $\displaystyle\int \sqrt{x^2+4x+13}\,dx$.

Showing the 12 most recent of 30 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.∫ (from 0 to 1) dx/√(1-x²) is equal to:

   (a) π/2

   (b) π/4

   (c) π/3

   (d) π/6

   ›Reveal solutionSolution

   $\frac{1}{\sqrt{1-x^2}}$ is the derivative of $\sin^{-1}x$, so the definite integral is just $\sin^{-1}(1) - \sin^{-1}(0)$.

   We know the standard result $\displaystyle\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}x + C$.

   So: …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2025Set ANNUAL1 markMCQ
   Q.If $\displaystyle\int_{\sqrt2}^{x} \dfrac{dt}{t\sqrt{t^2-1}} = \dfrac{\pi}{12}$, then the value of x is

   (a) 2

   (b) 1

   (c) 0

   (d) -1

   ›Reveal solutionSolution

   The antiderivative of $\dfrac{1}{t\sqrt{t^2-1}}$ is $\sec^{-1}t$; apply the limits and solve for $x$.

   We know the standard result:

   $\int\dfrac{dt}{t\sqrt{t^2-1}} = \sec^{-1}|t| + C$

   So

   $\int_{\sqrt2}^{x}\dfrac{dt}{t\sqrt{t^2-1}} = \sec^{-1}x - \sec^{-1}\sqrt2$

   Since $\sec\dfrac{\pi}{4} = \sqrt2$, we have $\sec^{-1}\sqrt2 = \dfrac{\pi}{4}$.

   Given this equals $\dfrac{\pi}{12}$:

   $\sec^{-1}x - \dfrac{\pi}{4} = \dfrac{\pi}{12}$ …

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2024Set A1 mark
   Q.Match the Correct Columns. Column A: $\int \sqrt{x^2-a^2}\,dx$. Column B options:

   (a) $\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+c$

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

   (c) $\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+c$

   (d) $\log\left|x+\sqrt{x^2-a^2}\right|+c$

   (e) $\dfrac{x}{2}\sqrt{x^2-a^2}-\dfrac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+c$

   (f) $\sec^{-1}x$

   (g) $\dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right|+c$

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

   (i) $\tan^{-1}x$. Which entry of Column B matches this Column A item?

   ›Reveal solutionSolution

   $\displaystyle\int\sqrt{x^2-a^2}\,dx$ matches entry (e), the standard NCERT formula for this integral.

   …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2024Set A1 mark
   Q.Match the Correct Columns. Column A: $\int \sqrt{a^2-x^2}\,dx$. Column B options:

   (a) $\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+c$

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

   (c) $\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+c$

   (d) $\log\left|x+\sqrt{x^2-a^2}\right|+c$

   (e) $\dfrac{x}{2}\sqrt{x^2-a^2}-\dfrac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+c$

   (f) $\sec^{-1}x$

   (g) $\dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right|+c$

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

   (i) $\tan^{-1}x$. Which entry of Column B matches this Column A item?

   ›Reveal solutionSolution

   $\displaystyle\int\sqrt{a^2-x^2}\,dx$ matches entry (c), the standard NCERT formula for this integral.

   …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2024Set A1 mark
   Q.Match the Correct Columns. Column A: $\int \dfrac{dx}{\sqrt{x^2-a^2}}$. Column B options:

   (a) $\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+c$

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

   (c) $\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+c$

   (d) $\log\left|x+\sqrt{x^2-a^2}\right|+c$

   (e) $\dfrac{x}{2}\sqrt{x^2-a^2}-\dfrac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+c$

   (f) $\sec^{-1}x$

   (g) $\dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right|+c$

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

   (i) $\tan^{-1}x$. Which entry of Column B matches this Column A item?

   ›Reveal solutionSolution

   $\displaystyle\int\dfrac{dx}{\sqrt{x^2-a^2}}$ matches entry (d).

   …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2024Set A1 mark
   Q.Match the Correct Columns. Column A: $\int \dfrac{dx}{\sqrt{a^2-x^2}}$. Column B options:

   (a) $\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+c$

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

   (c) $\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+c$

   (d) $\log\left|x+\sqrt{x^2-a^2}\right|+c$

   (e) $\dfrac{x}{2}\sqrt{x^2-a^2}-\dfrac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+c$

   (f) $\sec^{-1}x$

   (g) $\dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right|+c$

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

   (i) $\tan^{-1}x$. Which entry of Column B matches this Column A item?

   ›Reveal solutionSolution

   $\displaystyle\int\dfrac{dx}{\sqrt{a^2-x^2}}$ matches entry (b).

   …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2024Set A1 mark
   Q.Match the Correct Columns. Column A: $\int \dfrac{dx}{x^2-a^2}$. Column B options:

   (a) $\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+c$

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

   (c) $\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+c$

   (d) $\log\left|x+\sqrt{x^2-a^2}\right|+c$

   (e) $\dfrac{x}{2}\sqrt{x^2-a^2}-\dfrac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+c$

   (f) $\sec^{-1}x$

   (g) $\dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right|+c$

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

   (i) $\tan^{-1}x$. Which entry of Column B matches this Column A item?

   ›Reveal solutionSolution

   $\displaystyle\int\dfrac{dx}{x^2-a^2}$ matches entry (a).

   …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2024Set A1 mark
   Q.Match the Correct Columns. Column A: $\int \dfrac{dx}{a^2-x^2}$. Column B options:

   (a) $\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+c$

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

   (c) $\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+c$

   (d) $\log\left|x+\sqrt{x^2-a^2}\right|+c$

   (e) $\dfrac{x}{2}\sqrt{x^2-a^2}-\dfrac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+c$

   (f) $\sec^{-1}x$

   (g) $\dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right|+c$

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

   (i) $\tan^{-1}x$. Which entry of Column B matches this Column A item?

   ›Reveal solutionSolution

   $\displaystyle\int\dfrac{dx}{a^2-x^2}$ matches entry (g).

   …

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2023Set A1 mark
   Q.Match the following — find the correct match for $\int \tan x\,dx$ from:

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

   (b) $\cos 2x$

   (c) $\sec^2 x$

   (d) $2\cos 2x$

   (e) $\log|\text{cosec}\,x - \cot x| + c$

   (f) $-\log|\cos x| + c$

   (g) $\log|\sec x + \tan x| + c$

   ›Reveal solutionSolution

   $\displaystyle\int\tan x\,dx=-\log|\cos x|+c$, matching option (f).

   Write $\tan x=\dfrac{\sin x}{\cos x}$ and substitute $u=\cos x$, $du=-\sin x\,dx$: …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2023Set A1 mark
    Q.Match the following — find the correct match for $\int \cot x\,dx$ from:

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

    (b) $\cos 2x$

    (c) $\sec^2 x$

    (d) $2\cos 2x$

    (e) $\log|\text{cosec}\,x - \cot x| + c$

    (f) $-\log|\cos x| + c$

    (g) $\log|\sec x + \tan x| + c$

    ›Reveal solutionSolution

    $\displaystyle\int\cot x\,dx=\log|\sin x|+c$, matching option (a).

    Write $\cot x=\dfrac{\cos x}{\sin x}$ and substitute $u=\sin x$, $du=\cos x\,dx$: …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2023Set A1 mark
    Q.Match the following — find the correct match for $\int \sec x\,dx$ from:

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

    (b) $\cos 2x$

    (c) $\sec^2 x$

    (d) $2\cos 2x$

    (e) $\log|\text{cosec}\,x - \cot x| + c$

    (f) $-\log|\cos x| + c$

    (g) $\log|\sec x + \tan x| + c$

    ›Reveal solutionSolution

    $\displaystyle\int\sec x\,dx=\log|\sec x+\tan x|+c$, matching option (g). This is a standard result — a memorised integral, verified by differentiating the RHS.

    …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2023Set A1 mark
    Q.Match the following — find the correct match for $\int \text{cosec}\,x\,dx$ from:

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

    (b) $\cos 2x$

    (c) $\sec^2 x$

    (d) $2\cos 2x$

    (e) $\log|\text{cosec}\,x - \cot x| + c$

    (f) $-\log|\cos x| + c$

    (g) $\log|\sec x + \tan x| + c$

    ›Reveal solutionSolution

    $\displaystyle\int \text{cosec}\,x\,dx=\log|\text{cosec}\,x-\cot x|+c$, matching option (e). A standard integral, verified by differentiating the RHS.

    …

    🔒 Start your 14-day free trial to see the full solution →