Question

Integrate the function: $tan^{-1}x$

Answer 1

The correct answer is: $xtan^{-1}x-\frac{1}{2}log(1+x^2)+C$

Let $I=∫1.tan^{-1}x dx$

Taking $tan^{-1}x$ as first function and 1 as second function and integrating by parts,we
obtain

$I=tan^{-1}x∫1dx-∫[{(\frac{d}{dx}tan^{-1}x)∫1.dx}]dx$

$=tan^{-1}x.x-∫\frac{1}{1+x^2}.x dx$

$=xtan^{-1}x-\frac{1}{2}∫\frac{2x}{1+x^2} dx$

$=xtan^{-1}x-\frac{1}{2}log|1+x^2|+C$

$=xtan^{-1}x-\frac{1}{2}log(1+x^2)+C$