# How do you integrate int sqrtx e^x dx  using integration by parts?

Jun 10, 2018

$\sqrt{x} {e}^{x} - \frac{\sqrt{\pi} e r f i \left(\sqrt{x}\right)}{2} + C$

#### Explanation:

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

Integrate by parts: $\int f g p r i m e = f g - \int f p r i m e g$
$f = \sqrt{x}$, $g p r i m e = {e}^{x}$
$\downarrow$
$f p r i m e = \frac{1}{2 \sqrt{x}} , g = {e}^{x} :$
=$\sqrt{x} {e}^{x} - \int {e}^{x} / \left(2 \sqrt{x}\right) \mathrm{dx}$

Now solving:
$\int {e}^{x} / \left(2 \sqrt{x}\right) \mathrm{dx}$
Substitute $u = \sqrt{x} \to \mathrm{dx} = 2 \sqrt{x} \mathrm{du} :$
=$\frac{\sqrt{\pi}}{2} \int \frac{2 {e}^{{u}^{2}}}{\sqrt{\pi}} \mathrm{du}$

Now solving:
$\int \frac{2 {e}^{{u}^{2}}}{\sqrt{\pi}} \mathrm{du}$

This is a special integral, imaginary error function:
$= e r f i \left(u\right)$

Plug in solved integrals:
$\frac{\sqrt{\pi}}{2} \int \frac{2 {e}^{{u}^{2}}}{\sqrt{\pi}} \mathrm{du}$
=$\frac{\sqrt{\pi} e r f i \left(u\right)}{2}$

Undo substitution $u = \sqrt{x} :$
=$\frac{\sqrt{\pi} e r f i \left(\sqrt{x}\right)}{2}$

Plug in solved integrals:
$\sqrt{x} {e}^{x} - \int {e}^{x} / 2 \sqrt{x} \mathrm{dx}$
=$\sqrt{x} {e}^{x} - \frac{\sqrt{\pi} e r f i \left(\sqrt{x}\right)}{2}$

The problem is solved:
$\int \sqrt{x} {e}^{x} \mathrm{dx}$
=$\sqrt{x} {e}^{x} - \frac{\sqrt{\pi} e r f i \left(\sqrt{x}\right)}{2} + C$