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

Answer

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

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