Evaluate the following integrals:

Assume $x+\sin x=t$

$d(x+\sin x)=d t$

$(1+\cos x) d x=d t$

Substituting $\mathrm{t}$ and $\mathrm{dt}$ in given equation

$\Rightarrow \int \frac{\mathrm{dt}}{\mathrm{t}^{2}}$

$\Rightarrow \int \mathrm{t}^{-3} \mathrm{dt}$

$\Rightarrow \frac{t^{-2}}{-2}+c$

$\Rightarrow \frac{-1}{2 t^{2}}+c$

But $t=x+\sin x$

$\Rightarrow \frac{-1}{2(x+\sin x)^{2}}+c$