Analytical Integration of Arccot x

We can integrate  
\[cot^{-1} x\]
  by parts by writing  
\[cot^{-1}x=1 \times cot^{-1}x\]
.
Let  
\[u=cot^{-1}x \rightarrow cotu=x \rightarrow --cosec^2u \frac{du}{dx}=1\]
  then
\[\frac{du}{dx}=- \frac{1}{cosec^2u} =- \frac{1}{cot^2u+1}=-\frac{1}{x^2+1}\]
.
\[\frac{dv}{dx}=1 \rightarrow v=x\]

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

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