Question

Evaluate: ∫ (x tan^–1 x/ (1 + x²)^3/2) dx

Answer 1

Let x = tanθ

⇒ dx = sec²θ dθ

Then,

I = \(\int\limits \frac{tan\theta\,tan^{-1}(tan\theta)}{(1+tan^2\theta)^{3/2}}\)sec²θ dθ

= \(\int\frac{\theta\,tan \theta}{sec^3\theta}\)sec²θ dθ

= \(\int\frac{\theta\,tan \theta}{sec\theta}\)dθ

= \(\int \theta\,sin\theta\,d\theta\) 

= \(\theta\int sin\theta\,d\theta\) - \(\int(\frac{d\theta}{d\theta}\int sin\theta d\theta)d\theta +C\)

= - θ cosθ + sinθ + C

= \(\frac{-\theta}{\sqrt{1+tan^2\theta}}\) + \(\frac{tan\theta}{\sqrt{1+tan^2\theta}}\) + C

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

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