Question

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

Answer 1

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`