integral of arctan(1/x)

**laura35066** · April 19th 2010, 03:21 PM

Could someone please show me step by step the solution to the integral of arctan(1/x)?

**Soroban** · April 19th 2010, 03:43 PM

Hello, laura35066!

$I \;=\;\int \arctan\left(\frac{1}{x}\right)\,dx$

Integrate by parts . . .

. . $\begin{array}{ccccc}u &=& \arctan\left(x^{-1}\right) \\ \\[-3mm] du &=& \dfrac{\text{-}x^{-2}}{1 + (x^{-1})^2} &=& \dfrac{- dx}{1+x^2} \end{array} \qquad \begin{array}{ccc} dv &=& dx \\ \\ v &=& x \end{array}$

$I \;\;=\;\;x\arctan\left(\frac{1}{x}\right) + \int \frac{x\,dx}{1+x^2} \;\;=\;\;x\arctan\left(\frac{1}{x}\right) + \tfrac{1}{2}\ln\left|1 + x^2\right| + C$

**laura35066** · April 19th 2010, 03:54 PM

Thank you very much!

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