# How do you integrate int e^(sqrt(2x)) by parts?

Jan 19, 2017

$\int {e}^{\sqrt{2 x}} \mathrm{dx} = {e}^{\sqrt{2 x}} \left(\sqrt{2 x} - 1\right) + C$

#### Explanation:

$I = \int {e}^{\sqrt{2 x}} \mathrm{dx}$

Let $t = \sqrt{2 x}$. This implies that $\frac{1}{2} {t}^{2} = x$, which we differentiate to show that $\mathrm{dx} = t \textcolor{w h i t e}{.} \mathrm{dt}$. Then:

$I = \int {e}^{t} \left(t \textcolor{w h i t e}{.} \mathrm{dt}\right) = \int t {e}^{t} \mathrm{dt}$

We will use integration by parts now, which takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. For $\int t {e}^{t} \mathrm{dt}$, let:

$\left\{\begin{matrix}u = t \text{ "=>" "du=dt \\ dv=e^tdt" "=>" } v = {e}^{t}\end{matrix}\right.$

Then:

$I = u v - \int v \mathrm{du}$

$I = t {e}^{t} - \int {e}^{t} \mathrm{dt}$

$I = t {e}^{t} - {e}^{t} + C$

$I = {e}^{t} \left(t - 1\right) + C$

Returning to $x$ from $t = \sqrt{2 x}$:

$I = {e}^{\sqrt{2 x}} \left(\sqrt{2 x} - 1\right) + C$