Answer
$$\frac{x\sin 4x}{4}+\frac{1}{16}\cos4x+c .$$
Work Step by Step
We do the integration by parts; we choose $u= x$ and $dv=\cos 4x dx$. Then, $du=dx$, $v=\frac{1}{4}\sin 4x$, hence we have
$$\int x\cos 4x dx=\int udv=uv-\int vdu\\
=\frac{x\sin 4x}{4}-\frac{1}{4}\int \sin 4xdx\\
=\frac{x\sin 4x}{4}+\frac{1}{16}\cos4x+c .$$
