# Evaluate the following integrals:

Question:

Evaluate $\int \frac{1}{\mathrm{e}^{\mathrm{x}}+1} \mathrm{dx}$

Solution:

$\int \frac{1}{e^{x}+1} d x$

We can write above integral as

$\Rightarrow \int \frac{1+e^{x}-e^{x}}{e^{x}+1} d x$

Considering first integral:

$\int \frac{1+e^{x}}{1+e^{x}} d x$

Since the numerator and denominator are exactly same, our integrand simplifies to 1 and integrand becomes:

$\Rightarrow \int d x$

$\Rightarrow x$

$\therefore \int \frac{1+e^{x}}{1+e^{x}} d x=x \cdots$ (3)

Considering second integral:

$\int \frac{-e^{x}}{e^{x}+1} d x$

Let $u=1+e^{x}, d u=e^{x} d x$

Apply u - substitution:

$\int \frac{1}{\mathrm{u}}(-\mathrm{du})=-\ln |u|$

Replacing the value of $u$ we get,

$\int \frac{-e^{x}}{e^{x}+1} d x=-\ln \left|1+e^{x}\right|+C \cdots(4)$

From (3) and (4) we get,

$\Rightarrow \int \frac{1+e^{x}}{e^{x}+1} d x+\int \frac{-e^{x}}{e^{x}+1} d x=x-\ln \left|1+e^{x}\right|+C$

$\therefore \int \frac{1}{e^{x}+1} d x=x-\ln \left|1+e^{x}\right|+C$