What is the Integral of $\frac{{tan}^{- 1} x}{1 + x^{2}}$ $tan^-1x/(1+x^2)$ ?

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Answer 1 · 2016-08-03T13:38:46

$= \frac{1}{2} {({tan}^{- 1} x)}^{2} + C$ $= 1/2 (tan^-1 x )^2 + C$

$int \ tan^-1x/(1+x^2) \ dx$

$= int \ tan^-1 x * d/dx (tan^(-1) x) \ dx$

$= int d/dx (1/2(\ tan^-1 x )^2 ) \ dx$

$= 1/2 (tan^-1 x )^2 + C$

What is the Integral of tan^-1x/(1+x^2)tan−1x1+x2 tan^-1x/(1+x^2)?