Existence of improper integral.

$\sin x$ may be negative in $[1,\infty)$, so $\int_1^\infty\frac{\sin x}{x(1+x)}\,dx < \int_1^\infty\frac1{x^2}\,dx$ is not quite correct. What you can say, by the squeeze theorem, is that $$\left|\int_1^\infty\frac{\sin x}{x(1+x)}\,dx\right|\le\int_1^\infty\frac1{x(1+x)}\,dx\le\int_1^\infty\frac1{x^2}\,dx$$ and since the last integral converges, so does the leftmost integral without the absolute value signs.