Question 6 - A-Level Maths

Show that $\frac { 1 } { 5 \cosh x - 3 \sinh x } = \frac { \mathrm { e } ^ { x } } { m + \mathrm { e } ^ { 2 x } }$, where $m$ is an integer.
Use the substitution $u = \mathrm { e } ^ { x }$ to show that $$\int _ { 0 } ^ { \ln 2 } \frac { 1 } { 5 \cosh x - 3 \sinh x } \mathrm {~d} x = \frac { \pi } { 8 } - \frac { 1 } { 2 } \tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)$$