# How to evaluate this $\int_{-1}^1 (\frac{1}{x}+1-\frac{\sqrt{1-x^2}}{x})\arctan\frac{2}{x^2}dx$

I dont know how to evaluate this improper integral: $$I=\int_{-1}^1 \left(\frac{1}{x}+1-\frac{\sqrt{1-x^2}}{x}\right )\arctan\frac{2}{x^2}dx$$ At first I tried to do it in trigonometric substitution：$x=\sin t,dx=\cos tdt$ $$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\csc t+1-\cot t\right)\arctan(2\csc ^2t)\cos tdt$$ $$=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\cot t+\cos t-\cot t\cos t\right)\arctan(2\csc ^2t)dt$$ I don't know what to do next,any help will be appreciated.

• – lab bhattacharjee Sep 15 '17 at 5:14
• Thank you for the link, but that is a sum,but does that have anything to do with the integral? – JamesJ Sep 15 '17 at 5:19
• Write $$\arctan\dfrac2{x^2}=\arctan\dfrac1{x-1}-\arctan\dfrac1{x+1}$$ – lab bhattacharjee Sep 15 '17 at 5:22
• @,Thanks for your hint,I'll try trying. – JamesJ Sep 15 '17 at 5:38
• $\left(\frac{1}{x}-\frac{\sqrt{1-x^2}}{x}\right )\arctan\frac{2}{x^2}$ is an odd function and it's continue at $x=0$ – FDP Sep 15 '17 at 12:54

\begin{align} \mathcal{I} &=\int_{-1}^{1}\mathrm{d}x\,\left(\frac{1}{x}+1-\frac{\sqrt{1-x^{2}}}{x}\right)\arctan{\left(\frac{2}{x^{2}}\right)}\\ &=\int_{-1}^{0}\mathrm{d}x\,\left(\frac{1}{x}+1-\frac{\sqrt{1-x^{2}}}{x}\right)\arctan{\left(\frac{2}{x^{2}}\right)}\\ &~~~~~+\int_{0}^{1}\mathrm{d}x\,\left(\frac{1}{x}+1-\frac{\sqrt{1-x^{2}}}{x}\right)\arctan{\left(\frac{2}{x^{2}}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\left(-\frac{1}{x}+1+\frac{\sqrt{1-x^{2}}}{x}\right)\arctan{\left(\frac{2}{x^{2}}\right)};~~~\small{\left[x\mapsto-x\right]}\\ &~~~~~+\int_{0}^{1}\mathrm{d}x\,\left(\frac{1}{x}+1-\frac{\sqrt{1-x^{2}}}{x}\right)\arctan{\left(\frac{2}{x^{2}}\right)}\\ &=2\int_{0}^{1}\mathrm{d}x\,\arctan{\left(\frac{2}{x^{2}}\right)}.\\ \end{align}