The integral $\int_0^\infty\left|\frac{\sin x}{1+x}\right|\,dx$ diverges. In fact, we have
\begin{eqnarray}
\int_0^\infty\left|\frac{\sin x}{1+x}\right|\,dx&=&\sum_{k=0}^\infty\int_{k\pi}^{(k+1)\pi}\left|\frac{\sin x}{1+x}\right|\,dx=\sum_{k=0}^\infty\int_0^\pi\left|\frac{\sin(x+k\pi)}{1+k\pi+x}\right|\,dx\\
&=&\sum_{k=0}^\infty\int_0^\pi\left|\frac{\sin x}{1+k\pi+x}\right|\,dx
=\sum_{k=0}^\infty\int_0^\pi\frac{\sin x}{1+k\pi+x}\,dx\\
&\ge& \sum_{k=0}^\infty\int_0^\pi\frac{\sin x}{1+(k+1)\pi}\,dx=\sum_{k=0}^\infty\frac{2}{1+(k+1)\pi}=\infty.
\end{eqnarray}
Hence the integral $\int_0^\infty\frac{\sin x}{1+x}\,dx$ isn't absolutely convergent.In contrast we have
$$
\int_0^\infty\left|\frac{\cos x}{(1+x)^2}\right|\,dx\le\int_0^\infty\frac{1}{(1+x)^2}\,dx=1,
$$
i.e. the integral $\int_0^\infty\frac{\cos x}{(1+x)^2}\,dx$ is absolutely convergent.
However, the integral $\int_0^\infty\frac{\sin x}{1+x}\,dx$ does converge, because we have
$$
\int_0^\infty\frac{\sin x}{1+x}\,dx=\sum_{k=0}^\infty(-1)^k\int_0^\pi\frac{\sin x}{1+x+k\pi}\,dx=:\sum_{k=0}^\infty(-1)^ka_k,
$$
where the sequence $(a_k)$ satisfies the following:
\begin{eqnarray}
a_{k+1}-a_k&=&\int_0^\pi\sin x\left(\frac{1}{1+x+k\pi+\pi}-\frac{1}{1+x+k\pi}\right)\,dx\\
&=&-\int_0^\pi\frac{\pi\sin x}{(1+x+k\pi)(1+x+k\pi+\pi)}\,dx\le 0,
\end{eqnarray}
and
$$
0\le a_k\le \int_0^\pi\frac{\sin x}{1+k\pi}\,dx=\frac{2}{1+k\pi}\to 0.
$$
Notice that for every $\theta>0$ we have
$$
\int_0^\theta\frac{\sin x}{1+x}\,dx=-\frac{\cos x}{1+x}\Big|_0^\theta-\int_0^\theta\frac{\cos x}{(1+x)^2}\,dx=1-\frac{\cos\theta}{1+\theta}-\int_0^\theta\frac{\cos x}{(1+x)^2}\,dx.
$$
It follows that
$$
\int_0^\infty\frac{\sin x}{1+x}\,dx=1-\int_0^\infty\frac{\cos x}{(1+x)^2}\,dx.
$$
