Can someone please check that i am doing this correctly

$\displaystyle \int \ 5xe^{4x}$

$\displaystyle Let u = 5x $

$\displaystyle v= e^{4x}/4$

$\displaystyle du/dx = 5$

$\displaystyle dv/dx = e^{4x}$

So $\displaystyle 5x . e^{4x}/4 $ - $\displaystyle \int e^{4x}/4 . 5 dx$

then $\displaystyle \frac {\ 5xe^{4x}} {4} - \frac {5e^{4x}} {16}$

So factorised:

$\displaystyle \frac {\ 5e^{4x}} {4} $ x $\displaystyle (x - 1/4) + c $

Cheers