Using the definition \(\tanh y = \frac { \mathrm { e } ^ { y } - \mathrm { e } ^ { - y } } { \mathrm { e } ^ { y } + \mathrm { e } ^ { - y } }\), show that, for \(| x | < 1\),
$$\tanh ^ { - 1 } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)$$
Hence, or otherwise, show that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \tanh ^ { - 1 } x \right) = \frac { 1 } { 1 - x ^ { 2 } }\).
Use integration by parts to show that
$$\int _ { 0 } ^ { \frac { 1 } { 2 } } 4 \tanh ^ { - 1 } x \mathrm {~d} x = \ln \left( \frac { 3 ^ { m } } { 2 ^ { n } } \right)$$
where \(m\) and \(n\) are positive integers.