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

Mar 17, 2016

$\int x {e}^{\sqrt{x}} \mathrm{dx} = 2 {e}^{\sqrt{x}} \left({x}^{\frac{3}{2}} - 3 x + 6 \sqrt{x} - 6\right) + C$

#### Explanation:

First, we use substitution.

Let $t = \sqrt{x} \implies \mathrm{dt} = \frac{1}{2 \sqrt{x}} \mathrm{dx}$ and ${t}^{3} = {x}^{\frac{3}{2}}$

Then, substituting, we have

$\int x {e}^{\sqrt{x}} \mathrm{dx} = 2 \int {x}^{\frac{3}{2}} {e}^{\sqrt{x}} \frac{1}{2 \sqrt{x}} \mathrm{dx} = 2 \int {t}^{3} {e}^{t} \mathrm{dt}$

Next, we apply integration by parts three times, using the formula

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

Integration by Parts 1:

Let $u = {t}^{3}$ and $\mathrm{dv} = {e}^{t} \mathrm{dt}$
Then $\mathrm{du} = 3 {t}^{2} \mathrm{dt}$ and $v = {e}^{t}$

Applying the formula:
$2 \int {t}^{3} {e}^{t} \mathrm{dt} = 2 \left({t}^{3} {e}^{t} - 3 \int {t}^{2} {e}^{t} \mathrm{dt}\right)$

Integration by Parts 2:

Focusing on the remaining integral, let $u = {t}^{2}$ and $\mathrm{dv} = {e}^{t} \mathrm{dt}$
Then $\mathrm{du} = 2 t \mathrm{dt}$ and $v = {e}^{t}$

Applying the formula:
$\int {t}^{2} {e}^{t} \mathrm{dt} = {t}^{2} {e}^{t} - 2 \int t {e}^{t} \mathrm{dt}$

Integration by Parts 3:

Again, focusing on the remaining integral, let $u = t$ and $\mathrm{dv} = {e}^{t} \mathrm{dt}$
Then $\mathrm{du} = \mathrm{dt}$ and $v = {e}^{t}$

Applying the formula:
$\int t {e}^{t} \mathrm{dt} = t {e}^{t} - \int {e}^{t} \mathrm{dt} = t {e}^{t} - {e}^{t} + C$

Substituting our result into the second integration by parts step, we obtain

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

Substituting this into the first integration by parts step, we obtain

$2 \int {t}^{3} {e}^{t} \mathrm{dt} = 2 \left[{t}^{3} {e}^{t} - 3 {e}^{t} \left({t}^{2} - 2 t + 2\right)\right] + C$

$= 2 {e}^{t} \left({t}^{3} - 3 {t}^{2} + 6 t - 6\right) + C$

Finally, we substitute $t = \sqrt{x}$ back in to obtain our final result:

$\int x {e}^{\sqrt{x}} \mathrm{dx} = 2 {e}^{\sqrt{x}} \left({x}^{\frac{3}{2}} - 3 x + 6 \sqrt{x} - 6\right) + C$