How do I find #intxe^(2x)dx# using integration by parts?

1 Answer
Jan 26, 2015

Since #(fg)'=f'g+fg'#, by integrating one has that #fg=\int f'g+\int fg'#.

In your case, it is useful to see #f(x)=x#, and #g'(x)=e^{2x}#.
This means:
#f(x)=x \implies f'(x)=1#
#g'(x)=e^{2x} \implies g(x)=e^{2x}/2#

So we have that #\int fg'= fg- \int f'g#, that in our cases is

#\int xe^{2x} = xe^{2x}/2 - \int e^{2x}/2#, where the last integral is easily #e^{2x}/4#. So, we have that

#\int xe^{2x} = xe^{2x}/2 - e^{2x}/4#