# Intergration by parts

• June 12th 2009, 03:34 AM
jamm89
Intergration by parts
Can someone please check that i am doing this correctly
$\int \ 5xe^{4x}$
$Let u = 5x$
$v= e^{4x}/4$
$du/dx = 5$
$dv/dx = e^{4x}$
So $5x . e^{4x}/4$ - $\int e^{4x}/4 . 5 dx$
then $\frac {\ 5xe^{4x}} {4} - \frac {5e^{4x}} {16}$
So factorised:
$\frac {\ 5e^{4x}} {4}$ x $(x - 1/4) + c$
Cheers
• June 12th 2009, 03:46 AM
alexmahone
It is correct! (Clapping)

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