Question

Evaluate

${\int }_0^{\sqrt{3}}{tan}^{-1}\left(\frac{2x}{1-x^2}\right)dx$

Answer 1

$\begin{array}{c}{\int }_0^{\sqrt{3}}{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)dx\\ =x{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)-{\int }^{}\frac{x\left(\frac{2}{1-x^2}+\frac{4x^2}{{\left(1-x^2\right)}^2}\right)}{\frac{4x^2}{{\left(1-x^2\right)}^2}+1}dx\\ =xta{n}^{-1}\left(\frac{2x}{1-x^2}\right)-{\int }^{}\left(\frac{2x}{\left(1-x^2\right)\left(\frac{4x^2}{{\left(1-x^2\right)}^2}+1\right)}+\frac{4x^3}{{\left(1-x^2\right)}^2\left(\frac{4x^2}{{\left(1-x^2\right)}^2}+1\right)}\right)dx\\ =xta{n}^{-1}\left(\frac{2x}{1-x^2}\right)-\ln \left(x^2+1\right)+C\\ Applythelimit\\ =-\frac{3\ln \left(4\right)+\left(\sqrt{3}-3\right)\pi }{3}\end{array}$