Analytical Integration of Arctan x

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

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

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