Where did I go wrong in my integration (by parts)?

Question

Working:

$\int_{}^{}\frac{1}{4}x^2e^{\frac{x}{4}}dx$

$u = x^2, \frac{du}{dx} = 2x, \frac{dv}{dx} = e^{\frac{x}{4}}, v = 4e^{\frac{x}{4}}$

$\frac{1}{4}\left(4e^{\frac{x}{4}}x^2-\int \:4e^{\frac{x}{4}}2x\right)$

$\int \:4e^{\frac{x}{4\:}}2x:\:u=2x,\:\frac{du}{dx}=2,\:\frac{dv}{dx}=4e^{\frac{x}{4}},\:v=16e^{\frac{x}{4}}$

$\int \:4e^{\frac{x}{4\:}}2x\:=\:32xe^{\frac{x}{4}}-32e^{\frac{x}{4}}$

$\therefore \frac{1}{4}\int \:x^2e^{\frac{x}{4}}dx\:=\:\frac{1}{4}\left(4e^{\frac{x}{4}}x^2-32xe^{\frac{x}{4}}+32e^{\frac{x}{4}}\right)+c$

Answer 1

Starting from here: $\int \:4e^{\frac{x}{4\:}}2x:\:u=2x,\:\frac{du}{dx}=2,\:\frac{dv}{dx}=4e^{\frac{x}{4}},\:v=16e^{\frac{x}{4}}$

Note this next step that is different:

$\int \:4e^{\frac{x}{4\:}}2x\:=\:32xe^{\frac{x}{4}}-32\int e^{\frac{x}{4}}= \:32xe^{\frac{x}{4}}-128e^{\frac{x}{4}}+C$

You forgot to multiply by 4 when integrating $e^{x/4}$