Analytical Integration of Arcsech x

We can integrate  
\[sech^{-1} x\]
  by parts by writing  
\[sech^{-1}x=1 \times sech^{-1}x\]
.
Let  
\[u=sech^{-1}x \rightarrow sechu=x \rightarrow -sechutanhu \frac{du}{dx}=1\]
  then  
\[\frac{du}{dx}=- \frac{1}{sechutanhu} =- \frac{1}{x \sqrt{1-sech^2 u}}=-\frac{1}{x \sqrt{1-x^2}}\]
.
\[\frac{dv}{dx}=1 \rightarrow v=x\]

\[\begin{equation} \begin{aligned} \int 1 \times sech^{-1}xdx &= x sech^{-1}x - \int x \times - \frac{1}{x \sqrt{1-x^2}}dx \\ &= x sech^{-1}x+ \int \frac{1}{\sqrt{1-x^2}}dx \\ &= x sech^{-1}x sin^{-1}x+c \end{aligned} \end{equation}\]

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