Evaluate: $\int_{0}^{1} \dfrac{dx}{e^{x}+e^{-x}}$

> [!TLDR] > The integral simplifies by rewriting the denominator as $2\cosh x$, then substituting $t = e^{x}$ to get a rational function. The value is $\boxed{\arctan(e) - \frac{\pi}{4}}$. ## Why U‑Substitution Works Here The integrand $\frac{1}{e^{x}+e^{-x}}$ is a classic hyperbolic form — it’s actually $\frac{1}{2\cosh x}$. But the direct antiderivative of $\operatorname{sech} x$ isn’t something most of us remember. The trick is to notice that $e^{x}$ and $e^{-x}$ are reciprocals. That suggests a substitution that turns the sum into something algebraic: let $u = e^{x}$. Then $e^{-x} = 1/u$, and $dx = du/u$. The integral becomes a rational function in $u$, which we can handle with standard techniques. ## Step‑by‑Step Solution 1. **Rewrite the integrand** The denominator is $e^{x} + e^{-x}$. Multiply numerator and denominator by $e^{x}$ to get a cleaner form: $$ \frac{1}{e^{x}+e^{-x}} = \frac{e^{x}}{e^{2x}+1}. $$ This is optional but often makes the substitution more obvious. 2. **Substitute $u = e^{x}$** Then $du = e^{x}\,dx = u\,dx$, so $dx = \frac{du}{u}$. When $x = 0$, $u = e^{0} = 1$. When $x = 1$, $u = e^{1} = e$. The integral becomes: $$ \int_{0}^{1} \frac{dx}{e^{x}+e^{-x}} = \int_{1}^{e} \frac{1}{u + \frac{1}{u}} \cdot \frac{du}{u}. $$ 3. **Simplify the integrand** Inside the integral: $$ \frac{1}{u + \frac{1}{u}} \cdot \frac{1}{u} = \frac{1}{\frac{u^{2}+1}{u}} \cdot \frac{1}{u} = \frac{u}{u^{2}+1} \cdot \frac{1}{u} = \frac{1}{u^{2}+1}. $$ So the integral reduces to: $$ \int_{1}^{e} \frac{du}{u^{2}+1}. $$ 4. **Integrate** The antiderivative of $\frac{1}{u^{2}+1}$ is $\arctan u$. Therefore: …

Evaluate: ∫₀¹ dfracdxex+e-x

NCERT Exemplar · Q29

\int_{0}^{1} \dfrac{dx}{e^{x}+e^{-x}}

Puducherry CbseShort· 3mImportance★★★★★

Appeared in past exams:CBSE 2025· Set 65/4/1· 1mexactKCET 2018· Set A-1· 1mexact

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Concept understanding — U Substitution

U Substitution: The Reverse Chain Rule

The chain rule differentiates composite functions: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ — differentiate the outer function, then multiply by the derivative of the inside. Integration asks the reverse: given $\cos(x^2) \cdot 2x$ , find the original function. That's what u substitution does — it reverses the chain rule.

The Core Intuition

When an integral looks like "a function times the derivative of its inside," substitute the inside with $u$ and the derivative of the inside with $du$ . Consider:

\int 2x \cos(x^2) \, dx

Here $2x$ is the derivative of $x^2$ , and $x^2$ is the inside of $\cos(x^2)$ . Let $u = x^2$ , so $du = 2x \, dx$ :

\int \cos(u) \, du = \sin(u) + C = \sin(x^2) + C

Check: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ .

The Precise Statement

\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du \quad \text{where } u = g(x), \, du = g'(x) \, dx

Valid provided $g$ is differentiable and the resulting integral in $u$ is simpler.

The Step-by-Step Method

Identify a function $g(x)$ whose derivative $g'(x)$ also appears (possibly up to a constant factor).
Set $u = g(x)$ , compute $du = g'(x) \, dx$ .
Rewrite the entire integral in $u$ and $du$ — every $x$ and $dx$ must be replaced.
Integrate with respect to $u$ .
Substitute back $u = g(x)$ .

Watch out

You cannot mix variables. If any $x$ remains after substitution, you chose the wrong $u$ (or must solve for $x$ in terms of $u$ — rare).

A Second Example (with a constant factor)

Evaluate $\int x \sqrt{x^2 + 1} \, dx$ . Let $u = x^2 + 1$ , so $x \, dx = \frac{1}{2} du$ :

\int \sqrt{u} \cdot \tfrac{1}{2} du = \tfrac{1}{2} \cdot \tfrac{2}{3} u^{3/2} + C = \tfrac{1}{3} (x^2 + 1)^{3/2} + C

When Does It Work?

When the integrand is something times the derivative of something inside. Common patterns:

$x \cdot f(x^2)$ — derivative of $x^2$ is $2x$ , so $u = x^2$
$e^{g(x)} \cdot g'(x)$ — derivative of $g(x)$ appears
$\frac{g'(x)}{g(x)}$ — leads to $\log|g(x)|$ …

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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. The value is $\boxed{\arctan(e) - \frac{\pi}{4}}$ .

Why U‑Substitution Works Here

The integrand $\frac{1}{e^{x}+e^{-x}}$ is a classic hyperbolic form — it’s actually $\frac{1}{2\cosh x}$ . But the direct antiderivative of $\operatorname{sech} x$ isn’t something most of us remember. The trick is to notice that $e^{x}$ and $e^{-x}$ are reciprocals. That suggests a substitution that turns the sum into something algebraic: let $u = e^{x}$ . Then $e^{-x} = 1/u$ , and $dx = du/u$ . The integral becomes a rational function in $u$ , which we can handle with standard techniques.

Step‑by‑Step Solution

Rewrite the integrand The denominator is $e^{x} + e^{-x}$ . Multiply numerator and denominator by $e^{x}$ to get a cleaner form:

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

This is optional but often makes the substitution more obvious.

Substitute $u = e^{x}$ Then $du = e^{x}\,dx = u\,dx$ , so $dx = \frac{du}{u}$ . When $x = 0$ , $u = e^{0} = 1$ . When $x = 1$ , $u = e^{1} = e$ . The integral becomes:

\int_{0}^{1} \frac{dx}{e^{x}+e^{-x}} = \int_{1}^{e} \frac{1}{u + \frac{1}{u}} \cdot \frac{du}{u}.

Simplify the integrand Inside the integral:

\frac{1}{u + \frac{1}{u}} \cdot \frac{1}{u} = \frac{1}{\frac{u^{2}+1}{u}} \cdot \frac{1}{u} = \frac{u}{u^{2}+1} \cdot \frac{1}{u} = \frac{1}{u^{2}+1}.

So the integral reduces to:

\int_{1}^{e} \frac{du}{u^{2}+1}.

Integrate The antiderivative of $\frac{1}{u^{2}+1}$ is $\arctan u$ . Therefore: …

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Q.Evaluate: $\int_{0}^{1} \dfrac{dx}{e^{x}+e^{-x}}$

Concept understanding — U Substitution

U Substitution: The Reverse Chain Rule

The Core Intuition

The Precise Statement

The Step-by-Step Method

A Second Example (with a constant factor)

When Does It Work?

Why U‑Substitution Works Here

Step‑by‑Step Solution

Method: Substituting $u=e^{x}$ for $e^{x}\pm e^{-x}$ integrands

Steps

Common Mistakes

Q.Evaluate: ∫01dxex+e−x\int_{0}^{1} \dfrac{dx}{e^{x}+e^{-x}}∫01​ex+e−xdx​

Why U‑Substitution Works Here

Step‑by‑Step Solution

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Q.Evaluate: $\int_{0}^{1} \dfrac{dx}{e^{x}+e^{-x}}$

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