Starting from the definition of tanh in terms of exponentials, prove that \(\tanh ^ { - 1 } x = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)\). [ \(\beta\)
Given that \(y = \operatorname { tah } ^ { - 1 } \left( \frac { 1 - x } { 2 + x } \right) , \mathrm { s } \quad\) th \(\mathrm { t } ( 2 x + 1 ) \frac { \mathrm { dy } } { \mathrm { dx } } + 1 = 0\)
Hence find the first three terms in the Maclaurin's series for \(\tanh ^ { - 1 } \left( \frac { 1 - x } { 2 + x } \right)\) in the form
$$a \ln 3 + b x + c x ^ { 2 }$$
wh re \(a , b\) ad \(c\) are constants to be determined.