Using integration by parts to evaluate an integrals Can't understand how to solve this math: use integration by parts to evaluate this integrals:
$$\int x\sin(2x + 1) \,dx$$
can any one solve this so i can understand how to do this! Thanks :)

$\int uv'=uv-\int u'v$. 
Choose $u(x):=x$, $v'(x):=\sin(2x+1)$. Then $u'(x)=1$ and $v(x)=-\frac{\cos(2x+1)}{2}$. So
$$
\int x\sin(2x+1)\,dx=-x\frac{\cos(2x+1)}{2}+\int\frac{\cos(2x+1)}{2}\,dx=-x\frac{\cos(2x+1)}{2}+\frac{\sin(2x+1)}{4}+C.
$$
