Originally Posted by

**Nacho** $\displaystyle

\begin{gathered}

\int {\arctan \left( {\frac{1}

{x}} \right)dx} = x\arctan \left( {\frac{1}

{x}} \right) + \int {x \cdot \frac{1}

{{1 + \left( {\frac{1}

{x}} \right)^2 }}} \cdot \frac{1}

{{x^2 }}dx = \hfill \\

= x\arctan \left( {\frac{1}

{x}} \right) + \int {\frac{1}

{{x^2 + 1}}dx} = x\arctan \left( {\frac{1}

{x}} \right) + \arctan \left( x \right) + C \hfill \\

\end{gathered}

$

Whereat $\displaystyle u=\arctan \left(\dfrac{1}{x} \right)$ and $\displaystyle dv=1$