Complete a square after substitution and use a product derivative

Let $u=\sqrt x$, so $dx=2u\,du$. The integral becomes $\int\dfrac2{u^2+2u+2}du=\int\dfrac2{(u+1)^2+1}du=2\tan^{-1}(u+1)+C$. Thus the answer is $2\tan^{-1}(\sqrt x+1)+C$ for $x>0$.

The product rule gives $\dfrac d{dx}(x\sin x)=x\cos x+\sin x$.

Use $u=\ln x$ and $dv=(\sin x+x\cos x)dx$, so $du=dx/x$ and the preceding derivative gives $v=x\sin x$. The integral is $x\sin x\ln x-\int\sin x\,dx=x\sin x\ln x+\cos x+C$.