Analytical Integration of Arctanh x
\[tanh^{-1} x\]
by parts by writing \[tanh^{-1}x=1 \times tanh^{-1}x\]
.Let
\[u=tanh^{-1}x \rightarrow tanhu=x \rightarrow sech^2u \frac{du}{dx}=1\]
then\[\frac{du}{dx}= \frac{1}{sech^2u} =\frac{1}{1-tanh^2u}=\frac{1}{1-x^2}\]
.\[\frac{dv}{dx}=1 \rightarrow v=x\]
\[\begin{equation} \begin{aligned} \int 1 \times tanh^{-1}xdx &= x tanh^{-1}x - \int x \times \frac{1}{1-x^2}dx \\ &= xtanh^{-1}x+ \frac{1}{2} ln(1-x^2)+c \end{aligned} \end{equation}\]
