Q.Evaluate $\displaystyle\int \frac{dx}{x^2+16}$.

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:

   $\int_0^1 \frac{dx}{\sqrt{1-x^2}} = \Big[\sin^{-1}x\Big]_0^1 = \sin^{-1}(1) - \sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$

   ✓Final answer

   The value of the integral is $\frac{\pi}{2}$ (option a).

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 $\dfrac{\pi}{12}$:

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

   $\sec^{-1}x = \dfrac{\pi}{12}+\dfrac{\pi}{4} = \dfrac{\pi}{12}+\dfrac{3\pi}{12} = \dfrac{4\pi}{12} = \dfrac{\pi}{3}$

   So $x = \sec\dfrac{\pi}{3} = 2$.

   Check: $\sec^{-1}2-\sec^{-1}\sqrt2 = \dfrac{\pi}{3}-\dfrac{\pi}{4} = \dfrac{\pi}{12}$ ✓

   ✓Final answer

   $x = 2$ — option (a)

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.

   This is one of the standard "integral of $\sqrt{x^2-a^2}$" results, derived using the substitution $x=a\sec\theta$ and integration by parts. The standard formula is $\displaystyle\int\sqrt{x^2-a^2}\,dx = \dfrac{x}{2}\sqrt{x^2-a^2}-\dfrac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+c$, matching column-B entry (e).

   ✓Final answer

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

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.

   Using $x=a\sin\theta$ and integration by parts, the standard result is $\displaystyle\int\sqrt{a^2-x^2}\,dx = \dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+c$, matching column-B entry (c).

   ✓Final answer

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

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).

   This is a standard integral: with $x=a\sec\theta$, $\displaystyle\int\dfrac{dx}{\sqrt{x^2-a^2}} = \log\left|x+\sqrt{x^2-a^2}\right|+c$, matching column-B entry (d).

   ✓Final answer

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

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).

   This is a standard integral: with $x=a\sin\theta$, $\displaystyle\int\dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\dfrac{x}{a}+c$, matching column-B entry (b).

   ✓Final answer

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

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).

   Using partial fractions, $\dfrac{1}{x^2-a^2}=\dfrac{1}{2a}\left(\dfrac{1}{x-a}-\dfrac{1}{x+a}\right)$, so $\displaystyle\int\dfrac{dx}{x^2-a^2} = \dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+c$, matching column-B entry (a).

   ✓Final answer

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

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).

   Using partial fractions, $\dfrac{1}{a^2-x^2}=\dfrac{1}{2a}\left(\dfrac{1}{a-x}+\dfrac{1}{a+x}\right)$, so $\displaystyle\int\dfrac{dx}{a^2-x^2} = \dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right|+c$, matching column-B entry (g).

   ✓Final answer

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

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 $u=\cos x$, $du=-\sin x\,dx$:

   $\int\tan x\,dx=\int\frac{\sin x}{\cos x}\,dx=-\int\frac{du}{u}=-\log|u|+c=-\log|\cos x|+c.$

   ✓Final answer

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

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 $u=\sin x$, $du=\cos x\,dx$:

    $\int\cot x\,dx=\int\frac{\cos x}{\sin x}\,dx=\int\frac{du}{u}=\log|u|+c=\log|\sin x|+c.$

    ✓Final answer

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

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.

    Differentiating the claimed antiderivative: $\dfrac{d}{dx}\log|\sec x+\tan x|=\dfrac{\sec x\tan x+\sec^2x}{\sec x+\tan x}=\dfrac{\sec x(\tan x+\sec x)}{\sec x+\tan x}=\sec x$, confirming the result.

    ✓Final answer

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

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.

    Differentiating: $\dfrac{d}{dx}\log|\text{cosec}\,x-\cot x|=\dfrac{-\text{cosec}\,x\cot x+\text{cosec}^2x}{\text{cosec}\,x-\cot x}=\dfrac{\text{cosec}\,x(\text{cosec}\,x-\cot x)}{\text{cosec}\,x-\cot x}=\text{cosec}\,x$, confirming the result.

    ✓Final answer

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