# How do you integrate int sqrtx e^(x-1 ) dx  using integration by parts?

May 22, 2016

Please see the explanation section below.

#### Explanation:

First note that $\int \sqrt{x} {e}^{x - 1} \mathrm{dx} = \frac{1}{e} \int \sqrt{x} {e}^{x} \mathrm{dx}$

Let $t = \sqrt{x}$, so that $\mathrm{dt} = \frac{1}{2 \sqrt{x}} \mathrm{dx}$ and, therefore $\mathrm{dx} = 2 t$

Upon substitution, the integral becomes:

$\frac{1}{e} \int t {e}^{{t}^{2}} \left(2 t\right) \mathrm{dt}$

Now let $u = t$ and $\mathrm{dv} = {e}^{{t}^{2}} \left(2 t\right) \mathrm{dt}$

so that we get $\mathrm{du} = \mathrm{dt}$ and $v = {e}^{{t}^{2}}$ (integrate by substitution.

Apply the parts rule to get

$\frac{1}{e} \left[t {e}^{{t}^{2}} - \int {e}^{{t}^{2}} \mathrm{dt}\right]$

$\int {e}^{{t}^{2}} \mathrm{dt} = \frac{\sqrt{\pi}}{2} \text{erfi} \left(t\right)$ (the imaginary error function at $t$), so we get

$\frac{1}{e} \left[t {e}^{{t}^{2}} - \frac{\sqrt{\pi}}{2} \text{erfi} \left(t\right)\right] + C$

Reversing our substitution gets us:

$\int \sqrt{x} {e}^{x - 1} \mathrm{dx} = \frac{1}{e} \left[\sqrt{x} {e}^{\sqrt{x}} - \frac{\sqrt{\pi}}{2} \text{erfi} \left(\sqrt{x}\right)\right] + C$