How do you integrate ##[(e^(2x))sinx]dx##?

Integrate by parts twice using ##u = e^(2x)## both times.

After the second , you'll have

##int e^(2x)sinx dx = -e^(2x)cosx + 2e^(2x)sinx - 4 int e^(2x)sinx dx##

Note that the last integral is the same as the one we want. Call it ##I## for now.

##I = -e^(2x)cosx + 2e^(2x)sinx - 4 I##

So ##I = 1/5[-e^(2x)cosx + 2e^(2x)sinx] +C##

You may rewrite / simplify / factor as you see fit.