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 :)

I'm accustomed to seeing integration by parts expressed like this:
$$
\int u\,dv = uv - \int v\,du.
$$
The idea is that $u$ is something you can easily differentiate and $dv$ is something you can easily antidifferentiate and $\int v\,du$ is somehow more tractable than $\int u\,dv$.
$$
\int x\Big(\sin(2x+1)\,dx\Big) = \int x\,dv = xv - \int v\,dx.
$$
We have $dv = \sin(2x+1)$, so $v=\frac{-1}{2}\cos(2x+1)$.  So
$$
xv - \int v\,dx = \frac{-1}{2}x\cos(2x+1) - \int \frac{-1}{2} \cos(2x+1)\,dx.
$$
$$
=\frac{-1}{2}x\cos(2x+1) +\frac12  \int \cos(2x+1)\,dx
$$
$$
=\frac{-1}{2}x\cos(2x+1) +\frac14 \sin(2x+1) + C.
$$
