Analytical Integration of Arccoth x

We can integrate  
\[coth^{-1} x\]
  by parts by writing  
\[coth^{-1}x=1 \times coth^{-1}x\]
.
Let  
\[u=coth^{-1}x \rightarrow cothu=x \rightarrow -cosech^2u \frac{du}{dx}=1\]
  then
\[\frac{du}{dx}=- \frac{1}{cosech^2u} =-\frac{1}{cotnh^2u-1}=-\frac{1}{x^2-1}= \frac{1}{x^2-1}\]
.
\[\frac{dv}{dx}=1 \rightarrow v=x\]

\[\begin{equation} \begin{aligned} \int 1 \times coth^{-1}xdx &= x coth^{-1}x - \int x \times - \frac{1}{x^2-1}dx \\ &= xcoth^{-1}x- \int \frac{x}{x^2-1}dx \\ &= x cot^{-1}x - \frac{1}{2} ln(x^2-1) \end{aligned} \end{equation}\]

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