Here's what I tried:
$$ \int \frac{\cos^2x-\sin^2x}{\sin x(1+\sin^2 x)}dx = \int \frac{\cos^2x}{\sin x (1+\sin^2x)}dx - \int \frac{\sin^2x}{\sin x (1+\sin^2x)}dx $$ I tried then to divide with $ \sin x $ and $ \cos x $. Tried to use some trigonometric identities, but it didn't work, I just complicated it more.
And I can't see something that I can substitute.
Hint. One may write, with the change of variable $u=\cos x$, $$ \int \frac{\cos^2x}{\sin x (1+\sin^2x)}dx=\int \frac{\cos^2x\:\sin x}{\sin^2 x (1+\sin^2x)}dx=-\int \frac{u^2}{(1-u^2) (2-u^2)}du $$ and $$ \int \frac{\sin^2x}{\sin x (1+\sin^2x)}dx=\int \frac{\sin x}{ 1+\sin^2x}dx=-\int \frac{du}{2-u^2} $$ then the new integrals are easier to evaluate.