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

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Answer 1 · 2015-01-26T18:58:07

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

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

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