Integration by parts, two classics

Idea. The factor $x$ becomes $1$ when differentiated, simplifying the integral. The other factor ($e^x$ or $\sin x$) is easy to integrate.

First integral. With $u=x$ and $v'=e^x$ (so $u'=1$, $v=e^x$):
$$\int_{\ln2}^1xe^x\,dx=\bigl[xe^x\bigr]_{\ln2}^1-\int_{\ln2}^1e^x\,dx=\bigl[xe^x-e^x\bigr]_{\ln2}^1=0-(2\ln2-2)=2-2\ln2.$$

Second integral. With $u=x$ and $v'=\sin x$ (so $v=-\cos x$):
$$\int_0^\pi x\sin x\,dx=\bigl[-x\cos x\bigr]_0^\pi+\int_0^\pi\cos x\,dx=\pi+0=\pi.$$

Remark. The rule of thumb for which factor to differentiate is *LIATE*: logarithms and inverse functions first, then algebraic, then trigonometric, then exponential.