# How do you integrate int e^sqrtx by integration by parts method?

Sep 25, 2016

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

#### Explanation:

$\int {e}^{\sqrt{x}} \mathrm{dx}$

Let $t = \sqrt{x}$. This also implies that ${t}^{2} = x$, and this makes finding $\mathrm{dx}$ easier, in that we easily see that $2 t \mathrm{dt} = \mathrm{dx}$. Thus:

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

Now, using integration by parts, which takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. Let:

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

Thus:

$2 \int t {e}^{t} \mathrm{dt} = 2 \left[t {e}^{t} - \int {e}^{t} \mathrm{dt}\right] = 2 t {e}^{t} - 2 {e}^{t} + C$

Factoring and back-substituting with $t = \sqrt{x}$:

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