Evalute the following integrals: 1 e^x+1 dx - Vidyadip

Question

Evalute the following integrals:
$\int\frac{1}{\text{e}^\text{x}+1}\text{dx}$

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Answer

Let $\text{I}=\int\frac{1}{\text{e}^\text{x}+1}\text{dx}$ then,
$\text{I}=\int\frac{1}{\text{e}^\text{x}\Big[1+\frac{1}{\text{e}^\text{x}}\Big]}\text{dx}$
$\Rightarrow\text{I}=\int\frac{1}{\text{e}^\text{x}\big[1+\text{e}^{-\text{x}}\big]}\text{dx}\ .....\text{(i)}$
Let $1+\text{e}^{-\text{x}}=\text{t}$ then,
$\text{d}(1+\text{e}^{-\text{x}})=\text{dt}$
$\Rightarrow-\text{e}^{-\text{x}}\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{-\text{dt}}{\text{e}^{-\text{x}}}$
$\text{dx}=-\text{dt}\times\text{e}^\text{x}$
Putting $1 + e^{-x} = t$ and $dx = -e^x$ dt in equation (i), we get,
$\text{I}=\int\frac{1}{\text{e}^\text{x}\times\text{t}}\times-\text{e}^\text{x}\text{dt}$
$=-\int\frac{\text{dt}}{\text{t}}$
$=-\log|\text{t}|+\text{C}$
$=-\log|1+\text{e}^{-\text{x}}|+\text{C}$
$\therefore -\log|1+\text{e}^{-\text{x}}|+\text{C}$

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