# Question #f9b8d

Apr 16, 2017

$\ln | \frac{\sqrt{1 - {e}^{x}} - 1}{\sqrt{1 - {e}^{x}} + 1} | + C$

#### Explanation:

Let $u = \sqrt{1 - {e}^{x}} = {\left(1 - {e}^{x}\right)}^{\frac{1}{2}}$

By squaring both sides,

${u}^{2} = 1 - {e}^{x} R i g h t a r r o w - {e}^{x} = {u}^{2} - 1$

By differentiating $u$ w.r.t. $x$,

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2} {\left(1 - {e}^{x}\right)}^{- \frac{1}{2}} \cdot \left(- {e}^{x}\right) = \frac{- {e}^{x}}{2 \sqrt{1 - {e}^{x}}}$

By rewriting in terms of $u$,

$R i g h t a r r o w \frac{\mathrm{du}}{\mathrm{dx}} = \frac{{u}^{2} - 1}{2 u}$

By taking the reciprocal of both sdies,

$R i g h t a r r o w \frac{\mathrm{dx}}{\mathrm{du}} = \frac{2 u}{{u}^{2} - 1}$

By multiplying both sides by $\mathrm{du}$,

$R i g h t a r r o w \mathrm{dx} = \frac{2 u}{{u}^{2} - 1} \mathrm{du}$

Now, let us look at the integral in question.

$\int \frac{1}{\sqrt{1 - {e}^{x}}} \mathrm{dx}$

By the above substitution,

$= \int \frac{1}{\cancel{u}} \cdot \frac{2 \cancel{u}}{{u}^{2} - 1} \mathrm{du} = \int \frac{2}{{u}^{2} - 1} \mathrm{du}$

By the partial fraction: $\frac{2}{{u}^{2} - 1} = \frac{1}{u - 1} - \frac{1}{u + 1}$,

$= \int \left(\frac{1}{u - 1} - \frac{1}{u + 1}\right) \mathrm{du}$

By Log Rule,

$= \ln | u - 1 | - \ln | u + 1 | + C$

By the log property: $\ln x - \ln y = \ln \left(\frac{x}{y}\right)$,

$= \ln | \frac{u - 1}{u + 1} | + C$

By substituting back $u = \sqrt{1 - {e}^{x}}$,

$= \ln | \frac{\sqrt{1 - {e}^{x}} - 1}{\sqrt{1 - {e}^{x}} + 1} | + C$

I hope that this was clear.