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Integrate xe2x.

Posted: Wed Jul 13, 2022 8:08 pm
by answerhappygod
a) \(\frac{e^{2x}}{4} (x-\frac{1}{4})+C\)
b) \(\frac{e^{2x}}{4} (2x-1)+C\)
c) \(\frac{e^{2x}}{2} (2x-1)+C\)
d) \(\frac{e^{2x}}{4} (x+1)+C\)