Now, $tan \frac{x}{2} \; = \; tanh\frac{u}{2} \\ \therefore \frac{u}{2} \; = \; tanh^{-1} \Big( tan \frac{x}{2} \Big) \\ \; \\ \; \\ \therefore \frac{u}{2} \; = \; \frac{1}{2} \; log \; \bigg( \frac{1 + tan \frac{x}{2}}{ 1 - tan \frac{x}{2}} \bigg) \; \; \; \; \ldots \bigg( \because tanh^{-1}\theta = \frac{1}{2} log \Big( \frac{1 + \theta}{1 - \theta} \Big) \bigg) \\ \; \\ \; \\ \therefore u \; = \; log \Bigg( \frac{ tan\frac{\pi}{4} \; + \; tan\frac{x}{2} }{ 1 - tan\frac{\pi}{4} \; tan\frac{x}{2} } \Bigg) \; = \; log \; tan \Big( \frac{\pi}{4} + \frac{x}{2} \Big) \; \; \; \ldots \; \bigg( \because \; tan(A \; + \; B) \; = \; \frac{tanA \; + \; tanB}{1 \; - \; tanA \; tanB } \bigg) $