Practise integration by parts

Take $u=\ln x$, $dv/dx=x$, giving $du/dx=1/x$, $v=x^2/2$. The integral is $(x^2/2)\ln x-\frac12\int x\,dx=\frac{x^2}2\ln x-\frac{x^2}4+C$, for $x>0$.

Take $u=x$, $dv/dx=\cos x$, so $du/dx=1$, $v=\sin x$. The integral is $x\sin x-\int\sin x\,dx=x\sin x+\cos x+C$.

Take $u=x$, $dv/dx=e^{-x}$, so $du/dx=1$, $v=-e^{-x}$. The result is $-xe^{-x}-\int(-e^{-x})\,dx=-xe^{-x}-e^{-x}+C$.