How would you evaluate the following the indefinite integral? (use C for the constant integration)

$\int {e}^{x} \sqrt{\left(41 + {e}^{x}\right)} \mathrm{dx}$

May 12, 2018

$\int {e}^{x} \sqrt{41 + {e}^{x}} \mathrm{dx} = \frac{2}{3} {\left(41 + {e}^{x}\right)}^{\frac{3}{2}} + C$

Explanation:

This can be solved using a $u -$substitution.

Let $u = 41 + {e}^{x}$

Calculating its differential, we get

$\frac{\mathrm{du}}{\mathrm{dx}} = {e}^{x}$
$\mathrm{du} = {e}^{x} \mathrm{dx}$

And this differential $\mathrm{du}$ does show up in our integral $\int \textcolor{red}{{e}^{x}} \sqrt{41 + {e}^{x}} \textcolor{red}{\mathrm{dx}}$, so the substitution is valid.

Rewriting with the substitution, we get

$\int \sqrt{u} \mathrm{du} = \int {u}^{\frac{1}{2}} \mathrm{du} = \frac{2}{3} {u}^{\frac{3}{2}} + C$

Rewriting in terms of $x$, we get

$\int {e}^{x} \sqrt{41 + {e}^{x}} \mathrm{dx} = \frac{2}{3} {\left(41 + {e}^{x}\right)}^{\frac{3}{2}} + C$