I'm now practicing integration with Euler's Formula
$$\int{xe^{x}sin(x)dx}$$
$$ = Im\biggr[\int{xe^{x}e^{ix}}dx\biggr] = Im\biggr[\int{xe^{(1+i)x}dx\biggr]}$$
$$= Im\biggr[\frac{1}{1+i}\int{xd(e^{(1+i)x})}\biggr]$$
$$ = \frac{1}{2}Im\biggr[(1-i)\biggr(xe^{(1+i)x}-\int{e^{(1+i)x}dx}\biggr)\biggr] $$
correct me if i'm wrong, from line 2 to 3 this is integral by part
my question is in line 3 why we have
$$xe^{(1+i)x}-\int{e^{(1+i)x}dx} $$
instead of
$$xe^{(1+i)x}-\int{e^{(1+i)x}d(e^{(1+i)x})} $$
Thank You.
The integrate by parts formula is $$ \int u\;dv = uv - \int v\;du . \tag{1}$$ Here, we take $u=x, v=e^{(1+i)x}$ so that $du = dx$ and $dv = d\big( e^{(1+i)x}\big)$. Then put this into $(1)$ to get exactly $$ \int x\;d \big(e^{(1+i)x}\big) = xe^{(1+i)x} - \int e^{(i+i)x} dx $$