= Solution
Separate one factor of the <hyperbolic sine> and use the <hyperbolic Pythagorean identity> $\sinh^2x=\cosh^2x-1$. With $u=\cosh x$ and $du=\sinh x\,dx$, <integration by substitution> gives
$$
\int\frac{\sinh^3x}{\cosh^2x}\,dx
=\int\frac{u^2-1}{u^2}\,du
=\int(1-u^{-2})\,du
=\boxed{\cosh x+\frac1{\cosh x}+C}.
$$
For real $x$, the <hyperbolic cosine> never vanishes. Differentiating the result gives $\sinh x(1-\operatorname{sech}^2x)=\sinh^3x/\cosh^2x$, checking the sign of the reciprocal term.
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