As given:

$\tanh\dfrac{u}{2}=\tan\dfrac{x}{2}$

$\therefore \dfrac{u}{2}=\tanh^{-1}\Big({\tan\dfrac{x}{2}}\Big)$

$LHS: \dfrac{u}{2}$

Solving RHS, we get,

$\dfrac{1}{2}\log\Bigg( \dfrac {1+\tan{x\over2}}{{1-\tan{x\over2}}}\Bigg) $  using formula ${\tanh}^{-1} x= \dfrac{1}{2}\log\dfrac{(1+x)}{(1-x)} $

$\displaystyle =\dfrac{1}{2}\log\Bigg(\dfrac{{\tan\dfrac{\pi}{4}+\tan\dfrac{x}{2}}}{1-\tan\dfrac{\pi}{4}\tan\dfrac{x}{2}}\Bigg)$  $\because \tan \Big(\dfrac{\pi}{4}\Big)=1$

$=\dfrac{1}{2}\log \tan\Big({\dfrac{\pi}{4}+\dfrac{x}{2}}\Big)$    $\Big(using\ identity\ : \tan(a+b)= {\tan(a)+\tan(b)\over1-\tan(a)\tan(b)} \Big)$)

$\dfrac u2=\dfrac{1}{2}\log \tan\Big({\dfrac{\pi}{4}+\dfrac{x}{2}}\Big)$

Cancelling (1/2) on both sides,

$\therefore u=\log \tan \Big( \dfrac {\pi}{4} + \dfrac {x}{2} \Big)$

Hence Proved.