real analysis - Prob. 9, Chap. 6, in Baby Rudin: Which one of these two improper integrals converges absolutely and which one does not?

Here are the links to three earlier posts of mine on Prob. 9, Chap. 6, in Baby Rudin here on Math SE.

Prob. 9, Chap. 6, in Baby Rudin: Integration by parts for improper integrals

Prob. 9, Chap. 6, in Baby Rudin: Integration by parts for improper integrals with one infinite limit

Prob. 9, Chap. 6 in Baby Rudin: Integration by parts for an improper integral

Now my question is the following.

Which one of the integrals $\int_0^\infty \frac{ \cos x }{ 1+x } \ \mathrm{d} x$ and $\int_0^\infty \frac{\sin x}{ (1+x)^2 } \ \mathrm{d} x$ converges absolutely, and which one does not?

My Attempt:

For any $b > 0$, and for all $x \in [0, b]$, the following inequality holds:
$$ \left\lvert \frac{ \sin x }{ (1+x)^2 } \right\rvert \leq \frac{1}{(1+x)^2}, $$
which implies (by virtue of Theorem 6.12 (b) in Baby Rudin) that
$$ \int_0^b \left\lvert \frac{ \sin x }{ (1+x)^2 } \right\rvert \ \mathrm{d} x \leq \int_0^b \frac{1}{(1+x)^2} \ \mathrm{d} x = - \frac{1}{1+b} - \left( - \frac{1}{1+0} \right) = 1 - \frac{1}{1+b}; $$
moreover,
$$ \lim_{b \to \infty} \left( 1 - \frac{1}{1+b} \right) = 1. $$
So we can conclude that
$$ \int_0^\infty \left\lvert \frac{ \sin x }{ (1+x)^2 } \right\rvert \ \mathrm{d} x = \lim_{ b \to \infty} \int_0^b \left\lvert \frac{ \sin x }{ (1+x)^2 } \right\rvert \ \mathrm{d} x \leq 1. $$

That is, the improper integral $\int_0^\infty \left\lvert \frac{ \sin x }{ (1+x)^2 } \right\rvert \ \mathrm{d} x $ converges, which is the same as saying that the integral $\int_0^\infty \frac{ \sin x }{ (1+x)^2 } \ \mathrm{d} x $ converges absolutely.

Am I right?

If so, then, as suggested by Rudin, the other integral does not converge absolutely.

But how to show this directly?