Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 17 / a

Use integration by substitution with $u=\ln x$, so $du=dx/x$. For $x>1$, also $u>0$, andThe domain condition explains why no absolute value is needed around the inner logarithm.

Separate one factor of the hyperbolic sine and use the hyperbolic Pythagorean identity ${\sinh }^{2}x={\cosh }^{2}x-1$. With $u=\cosh x$ and $du=\sinh xdx$, integration by substitution givesFor real $x$, the hyperbolic cosine never vanishes. Differentiating the result gives $\sinh x(1-{\mathrm{sech}}^{2}x)={\sinh }^{3}x/{\cosh }^{2}x$, checking the sign of the reciprocal term.