1. ## intergration by parts

Hi

So I am solving integral sign (S) (x-1)e^-2xdx by using integration by parts

u = x-1 dv = e^-2x dx
du = dx v= (-1/2)e^-2x

so S udv = uv - S vdu

so = (x-1)((-1/2)e^-2x) - S (-1/2)e^-2x dx

Now my question is when I solve the S dx part does it become x-1 or just x?

Thanks

Calculus beginner

2. I'm not sure I understand your question. Your calculations are correct. Just solve the remaining integral $\frac{-1}{2}\int e^{-2x}dx=\frac{1}{4}e^{-2x}$.