What is the integration of x tan inverse x dx ?

Solution :

Let I = \(\int\) x\(tan^{-1}x\) dx

By using Integration by parts rule,

Taking tan inverse x as first function and x as second function. Then,

I = (\(tan^{-1}x\)) \(\int\) x dx – \(\int\){\({d\over dx}\)(\(tan^{-1}x\) \(\int\) x dx} dx

I = (\(tan^{-1}x\))\(x^2\over 2\) – \(\int\)\({1\over 1 + x^2}\) \(\times\) \(x^2\over 2\) dx

\(\implies\) I = \(x^2\over 2\)\(tan^{-1}x\) – \(1\over 2\) \(\int\) \(x^2 + 1 – 1\over x^2 + 1\) dx

\(\implies\) I = \(x^2\over 2\)\(tan^{-1}x\) – \(1\over 2\) \(\int\) 1 – \({1\over x^2 + 1}\) dx

\(\implies\) I = \(x^2\over 2\)\(tan^{-1}x\) – \(1\over 2\) (\(x – tan^{-1}x\)) + C