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

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

\(=tan^{-1}x.x-∫\frac{1}{1+x^2}.x dx\)

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

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

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