Consider the integral ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}4x{e}^{\frac{x}{4}}dx$ :

Applying the integration by parts technique, let

$u={}^6$ and

$dv=e^{\frac{x}{4}},{}^{6}dx$

Then $uv-{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}$vdu$=$

$$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}dx$

So ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}4x{e}^{\frac{}{4}}dx=$