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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals4 Marks
Question
Evalute the following integrals:
$\int\frac{1}{\text{e}^\text{x}+1}\text{dx}$
✓

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 = -ex 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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