Finding an antiderivative of $\sin^2x\over (1+x)^2$ would not be easy, so we will use the comparison test for integrals with unbounded regions of integration. Since $\sin^2 x$ is nonnegative and bounded by $1$, $0\le { \sin^2x\over(1+x)^2}\le {1\over(1+x)^2}.$ The given integral converges if the integral $ \int_0^\infty {1\over(1+x)^2}\,dx $ converges.
Now $ \eqalign{ \int {1\over(1+x)^2}\, dx\buildrel{u=1+x}\over{ =}\int {1\over u^2} \, du={-1 \over 1+x}+C. } $
So $\eqalign{ \int_0^\infty {1\over(1+x)^2}\, dx& =\lim_{b\rightarrow\infty}\int_0^b {1\over(1+x)^2}\, dx\cr &=\lim_{b\rightarrow\infty} {-1 \over 1+x}\Bigl|_0^b\cr &=0-(-1)\cr &=1. } $
Thus $\int_0^\infty {1\over(1+x)^2}\, dx$ converges; and so, as mentioned above, $\int_0^\infty {\sin^2 x\over(1+x)^2}\, dx$ converges.