The value of $\int_0^1 \frac{1}{e^x + e^{-x}}\, dx$ is : (A) $-\frac{\pi}{4}$ (B) $\frac{\pi}{4}$ (C) $\tan^{-1} e - \frac{\pi}{4}$ (D) $\tan^{-1} e$

> [!TLDR] > The integral simplifies by rewriting the denominator as $2\cosh x$, then substituting $t = e^x$ to get a rational function in $t$, which integrates to an arctangent. The value is $\tan^{-1} e - \frac{\pi}{4}$, which is option **(C)**. The key insight here is that $e^x + e^{-x}$ is exactly $2\cosh x$, but more usefully, it suggests a substitution that turns the integral into a standard arctangent form. When you see a sum of exponentials in the denominator, your first instinct should be to multiply numerator and denominator by something to simplify — here, multiplying by $e^x$ does the trick. Let’s work through it. 1. **Rewrite the integrand** Multiply numerator and denominator by $e^x$: \[ \frac{1}{e^x + e^{-x}} = \frac{e^x}{e^{2x} + 1}. \] This is cleaner because the denominator is now $e^{2x}+1$, which looks like $u^2+1$ after a substitution. 2. **Substitute $t = e^x$** Then $dt = e^x\,dx$, so $dx = \frac{dt}{t}$. But notice: the numerator already has $e^x\,dx$ in disguise. When $x = 0$, $t = e^0 = 1$. When $x = 1$, $t = e^1 = e$. The integral becomes: \[ \int_0^1 \frac{e^x}{e^{2x}+1}\,dx = \int_{1}^{e} \frac{1}{t^2+1}\,dt. \] That’s a direct substitution — no extra factor needed because $e^x dx = dt$. 3. **Integrate the arctangent form** The integral $\int \frac{1}{t^2+1}\,dt$ is $\tan^{-1} t + C$. So: \[ \int_{1}^{e} \frac{1}{t^2+1}\,dt = \left[ \tan^{-1} t \right]_{1}^{e} = \tan^{-1} e - \tan^{-1} 1. \] 4. **Evaluate the known arctangent** $\tan^{-1} 1 = \frac{\pi}{4}$. Therefore: \[ …

The value of ∫₀¹ frac1ex + e-x dx is : (A) -(π)/(4) (B) — CBSE Class 12 Mathematics 2025 PYQ

Question

\int_0^1 \frac{1}{e^x + e^{-x}}\, dx

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The integral simplifies by rewriting the denominator as $2\cosh x$ , then substituting $t = e^x$ to get a rational function in $t$ , which integrates to an arctangent. The value is $\tan^{-1} e - \frac{\pi}{4}$ , which is option (C).

The key insight here is that $e^x + e^{-x}$ is exactly $2\cosh x$ , but more usefully, it suggests a substitution that turns the integral into a standard arctangent form. When you see a sum of exponentials in the denominator, your first instinct should be to multiply numerator and denominator by something to simplify — here, multiplying by $e^x$ does the trick.

Let’s work through it.

Rewrite the integrand Multiply numerator and denominator by $e^x$ :

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

This is cleaner because the denominator is now $e^{2x}+1$ , which looks like $u^2+1$ after a substitution.

Substitute $t = e^x$ Then $dt = e^x\,dx$ , so $dx = \frac{dt}{t}$ . But notice: the numerator already has $e^x\,dx$ in disguise. When $x = 0$ , $t = e^0 = 1$ . When $x = 1$ , $t = e^1 = e$ . The integral becomes:

\int_0^1 \frac{e^x}{e^{2x}+1}\,dx = \int_{1}^{e} \frac{1}{t^2+1}\,dt.

That’s a direct substitution — no extra factor needed because $e^x dx = dt$ .

Integrate the arctangent form The integral $\int \frac{1}{t^2+1}\,dt$ is $\tan^{-1} t + C$ . So:

\int_{1}^{e} \frac{1}{t^2+1}\,dt = \left[ \tan^{-1} t \right]_{1}^{e} = \tan^{-1} e - \tan^{-1} 1.

Evaluate the known arctangent $\tan^{-1} 1 = \frac{\pi}{4}$ . Therefore:

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 29 on this concept.

CBSE 2026Set 65/2/11 markMCQ
$\int_{0}^{2a} \frac{1}{1+4x^2}\, dx = \frac{\pi}{6}$
›Reveal solutionSolution
The key idea is to evaluate the definite integral using the standard arctangent formula, set it equal to $\frac{\pi}{6}$ , and solve for $a$ . The value of $a$ is $\frac{\sqrt{3}}{4}$ .

We are given:
$\int_{0}^{2a} \frac{1}{1+4x^2}\, dx = \frac{\pi}{6}$
and need to find $a$ from the options.

Concept and intuition:

The integrand $\frac{1}{1+4x^2}$ looks like the derivative of an inverse trigonometric function. Recall that $\frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2}$ . Here, the denominator has $4x^2$ instead of $x^2$ , so a substitution or a standard formula adjustment is needed. The standard result is:
$\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C$
Our denominator is $1 + 4x^2 = 1 + (2x)^2$ , so we can treat it as $1^2 + (2x)^2$ . This suggests letting $u = 2x$ , which will convert the integral into the standard arctangent form.

Let’s work through it step by step.

Rewrite the integral in a standard form. The denominator is $1 + 4x^2 = 1 + (2x)^2$ . So we have:

$\int \frac{dx}{1 + (2x)^2}$
This matches $\int \frac{dx}{a^2 + u^2}$ with $a = 1$ and $u = 2x$ , but we need to account for the $dx$ vs $du$ change.

Substitute $u = 2x$ . Then $du = 2\, dx$ , so $dx = \frac{du}{2}$ . The limits: when $x = 0$ , $u = 0$ ; when $x = 2a$ , $u = 4a$ . The integral becomes:

$\int_{0}^{2a} \frac{1}{1 + (2x)^2}\, dx = \int_{u=0}^{u=4a} \frac{1}{1 + u^2} \cdot \frac{du}{2} = \frac{1}{2} \int_{0}^{4a} \frac{du}{1+u^2}$

Evaluate the standard integral. We know $\int \frac{du}{1+u^2} = \tan^{-1} u + C$ . So:

$\frac{1}{2} \int_{0}^{4a} \frac{du}{1+u^2} = \frac{1}{2} \left[ \tan^{-1} u \right]_{0}^{4a} = \frac{1}{2} \left( \tan^{-1}(4a) - \tan^{-1}(0) \right)$
Since $\tan^{-1}(0) = 0$ , this simplifies to:
$\frac{1}{2} \tan^{-1}(4a)$

Set equal to the given value and solve. The problem states this equals $\frac{\pi}{6}$ :

$\frac{1}{2} \tan^{-1}(4a) = \frac{\pi}{6}$
Multiply both sides by 2:
$\tan^{-1}(4a) = \frac{\pi}{3}$
Now take the tangent of both sides:
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CBSE 2026Set A1 markMCQ
$\int_0^1\frac{dx}{1 + x^2} =$
(a) $\frac{\pi}{2}$
(b) $\frac{\pi}{3}$
(c) $\frac{\pi}{4}$
(d) $\pi$
›Reveal solutionSolution

$\int_0^1\tfrac{dx}{1+x^2}=\tan^{-1}1-\tan^{-1}0=\tfrac{\pi}{4}$ .

Since $\int\dfrac{dx}{1+x^2}=\tan^{-1}x$ , evaluate between the limits:

…

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CBSE 2026Set A1 markMCQ
$\int_0^1\frac{x^3\,dx}{1 + x^8} =$
(a) $\frac{\pi}{2}$
(b) $\frac{\pi}{4}$
(c) $\frac{\pi}{8}$
(d) $\frac{\pi}{16}$
›Reveal solutionSolution

Substitute $t=x^4$ : $\int_0^1\tfrac{x^3\,dx}{1+x^8}=\tfrac14\int_0^1\tfrac{dt}{1+t^2}=\tfrac{\pi}{16}$ .

Let $t=x^4$ , so $dt=4x^3\,dx$ , i.e. $x^3\,dx=\tfrac{dt}{4}$ . Limits: $x=0\to t=0$ , $x=1\to t=1$ . Also $x^8=t^2$ . Then

…

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CBSE 2026Set ANNUAL1 markMCQ
$\int \dfrac{dx}{x^2+16} =$
(a) $\dfrac{1}{4}\tan^{-1}\left(\dfrac{x}{4}\right)+C$
(b) $\tan^{-1} x + C$
(c) $\dfrac{1}{2}\tan^{-1}\left(\dfrac{x}{2}\right)+C$
(d) None of these
›Reveal solutionSolution

Use the standard integral $\int \dfrac{dx}{x^2+a^2} = \dfrac{1}{a}\tan^{-1}\left(\dfrac{x}{a}\right)+C$ .

Here $a^2=16 \Rightarrow a=4$ .

…

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CBSE 2026Set ANNUAL1 mark
$\displaystyle\int_0^1 \dfrac{dx}{1+x^2}$
›Reveal solutionSolution

Use the standard integral $\int\frac{dx}{1+x^2}=\tan^{-1}x+C$ and apply the limits.

$\int_0^1 \frac{dx}{1+x^2} = \Big[\tan^{-1}x\Big]_0^1$

…

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CBSE 2026Set ANNUAL1 mark
$\displaystyle\int_{0}^{a}\dfrac{dx}{1+4x^{2}}=\dfrac{\pi}{8}$
›Reveal solutionSolution

$\int_0^a\frac{dx}{1+4x^2}=\tfrac12\tan^{-1}(2a)=\tfrac{\pi}{8}$ gives $2a=1$ , i.e. $a=\tfrac12$ .

Step 1: $\displaystyle\int\frac{dx}{1+(2x)^{2}}=\frac{1}{2}\tan^{-1}(2x)$ .

Step 2: $\displaystyle\int_{0}^{a}\frac{dx}{1+4x^{2}}=\frac{1}{2}\tan^{-1}(2a)-0=\frac{1}{2}\tan^{-1}(2a).$

…

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CBSE 2025Set 65/4/11 markMCQ
$\int_0^1 \frac{1}{e^x + e^{-x}}\, dx$
›Reveal solutionSolution
The integral simplifies by rewriting the denominator as $2\cosh x$ , then substituting $t = e^x$ to get a rational function in $t$ , which integrates to an arctangent. The value is $\tan^{-1} e - \frac{\pi}{4}$ , which is option (C).

The key insight here is that $e^x + e^{-x}$ is exactly $2\cosh x$ , but more usefully, it suggests a substitution that turns the integral into a standard arctangent form. When you see a sum of exponentials in the denominator, your first instinct should be to multiply numerator and denominator by something to simplify — here, multiplying by $e^x$ does the trick.

Let’s work through it.

Rewrite the integrand Multiply numerator and denominator by $e^x$ :

$\frac{1}{e^x + e^{-x}} = \frac{e^x}{e^{2x} + 1}.$
This is cleaner because the denominator is now $e^{2x}+1$ , which looks like $u^2+1$ after a substitution.

Substitute $t = e^x$ Then $dt = e^x\,dx$ , so $dx = \frac{dt}{t}$ . But notice: the numerator already has $e^x\,dx$ in disguise. When $x = 0$ , $t = e^0 = 1$ . When $x = 1$ , $t = e^1 = e$ . The integral becomes:

$\int_0^1 \frac{e^x}{e^{2x}+1}\,dx = \int_{1}^{e} \frac{1}{t^2+1}\,dt.$
That’s a direct substitution — no extra factor needed because $e^x dx = dt$ .

Integrate the arctangent form The integral $\int \frac{1}{t^2+1}\,dt$ is $\tan^{-1} t + C$ . So:

$\int_{1}^{e} \frac{1}{t^2+1}\,dt = \left[ \tan^{-1} t \right]_{1}^{e} = \tan^{-1} e - \tan^{-1} 1.$

Evaluate the known arctangent $\tan^{-1} 1 = \frac{\pi}{4}$ . Therefore:
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CBSE 2025Set E1 markMCQ
$\int \frac{dx}{x^2 + 4} =$
(a) $\frac{1}{4}\tan^{-1}\frac{x}{4} + k$
(b) $\frac{1}{2}\tan^{-1}\frac{x}{2} + k$
(c) $\frac{1}{2}\tan^{-1}\frac{2}{x} + k$
(d) $2\tan^{-1}\frac{x}{2} + k$
›Reveal solutionSolution

Apply the standard integral $\int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\frac{x}{a}+k$ with $a=2$ .

Here $x^2+4=x^2+2^2$ , so $a=2$ : …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. \int (from 0 to 1) dx / (1 + x²) is equal to:$
(a) π/2
(b) π/4
(c) π/3
(d) π/6
›Reveal solutionSolution

$\int \dfrac{dx}{1+x^2} = \tan^{-1}x + C$ ; evaluate at the limits.

…

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CBSE 2024Set ANNUAL1 mark
$\displaystyle\int \dfrac{1}{x^2+36}\, dx$
›Reveal solutionSolution

Apply the standard formula $\int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\frac xa+C$ with $a=6$ .

Here $x^2+36 = x^2+6^2$ , so $a=6$ . Using the standard integral:

$\int \frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C$

…

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CBSE 2024Set D1 markMCQ
$\int \frac{dx}{a^2 + x^2} =$
(a) $\frac{1}{a}\tan^{-1}\frac{a}{x} + c$
(b) $\tan^{-1}\frac{x}{a} + c$
(c) $\frac{1}{a}\tan^{-1}\frac{x}{a} + c$
(d) $\frac{1}{a}\tan^{-1} x + c$
›Reveal solutionSolution

Standard form: $\int\dfrac{dx}{a^2+x^2}=\dfrac1a\tan^{-1}\dfrac{x}{a}+c$ .

This is a memorised standard result:

…

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CBSE 2024Set D1 markMCQ
$\int \frac{x^3\,dx}{1 + x^8} =$
(a) $\tan^{-1} x^4 + c$
(b) $4\tan^{-1} x^4 + c$
(c) $\frac{1}{4}\tan^{-1} x^4 + c$
(d) $2\tan^{-1} x^4 + c$
›Reveal solutionSolution

Substitute $u = x^4$ so the integral becomes a standard $\tan^{-1}$ form.

Let $u = x^4 \Rightarrow du = 4x^3\,dx$ , so $x^3\,dx = \frac{1}{4}du$ . Then …

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Q.The value of $\int_0^1 \frac{1}{e^x + e^{-x}}\, dx$ is : (A) $-\frac{\pi}{4}$ (B) $\frac{\pi}{4}$ (C) $\tan^{-1} e - \frac{\pi}{4}$ (D) $\tan^{-1} e$

Concept understanding — Definite Integral Arctangent

Definite Integral of $\dfrac{1}{1+x^2}$

The core idea

A worked value

Q.The value of ∫011ex+e−x dx\int_0^1 \frac{1}{e^x + e^{-x}}\, dx∫01​ex+e−x1​dx is : (A) −π4-\frac{\pi}{4}−4π​ (B) π4\frac{\pi}{4}4π​ (C) tan⁡−1e−π4\tan^{-1} e - \frac{\pi}{4}tan−1e−4π​ (D) tan⁡−1e\tan^{-1} etan−1e

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Q.The value of $\int_0^1 \frac{1}{e^x + e^{-x}}\, dx$ is : (A) $-\frac{\pi}{4}$ (B) $\frac{\pi}{4}$ (C) $\tan^{-1} e - \frac{\pi}{4}$ (D) $\tan^{-1} e$

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