# $\int_{0}^{\infty}\tan^{-1}\left(\frac{2x}{1+x^2}\right)\frac{x}{x^2+4}dx$

$$\int_{0}^{\infty}\tan^{-1}\left(\frac{2x}{1+x^2}\right)\frac{x}{x^2+4}dx$$

I am given solution for this definite integral is $$\frac{\pi}{2}\left(\ln\frac{\sqrt2+3}{\sqrt2+1}\right)$$. Any idea or approach you would use to solve this?

\begin{align} &\int_{0}^{\infty}\tan^{-1}\left(\frac{2x}{1+x^2}\right)\frac{x}{x^2+4}\ dx\\ =& \int_{0}^{\infty}\int_{\sqrt2-1}^{\sqrt2+1}\frac{x}{x^2+y^2}\frac{x}{x^2+4}\ dy \ dx\\ =& \ \frac\pi2 \int_{\sqrt2-1}^{\sqrt2+1}\frac1{y+2}dy= \frac{\pi}{2}\ln\frac{\sqrt2+3}{\sqrt2+1} \end{align}

• A wonderful solutio indeed! Sep 12 at 4:33
• Wow this is elegant. Sep 26 at 15:52

Using integration by parts, we find $$\int_0^\infty \tan^{-1}\left(\frac{2 x} {x^2 +1}\right)\frac{x}{x^2 +4}dx=\int_0^{\infty}\frac{(x^2-1)\ln(x^2+4)}{x^4+6x^2+1} dx.$$ Factoring the denominator as $$x^4+6x^2+1=(x^2+3+2\sqrt{2})(x^2+3-2\sqrt{2})\\ =(x^2+(\sqrt{2}+1)^2)(x^2+(\sqrt{2}-1)^2)$$ and then applying the partial fraction decomposition, we obtain $$\frac{\sqrt{2}+1}{2}\int_0^{\infty} \frac{\ln(x^2+4)}{x^2+(\sqrt{2}+1)^2} dx - \frac{\sqrt{2}-1}{2}\int_0^{\infty} \frac{\ln(x^2+4)}{x^2+(\sqrt{2}-1)^2} dx.$$ Finally we use An improper integral : $\int_{0}^\infty {\ln(a^2+x^2)\over{b^2+x^2}}dx$? $$=\frac{\pi}{b} \, \ln(a+b)$$, getting $$\frac{\pi}{2}\ln(\sqrt{2}+3)-\frac{\pi}{2}\ln(\sqrt{2}+1)=\frac{\pi}{2}\ln\left(\frac{\sqrt2+3}{\sqrt2+1}\right).$$

Scientia, Series A: Mathematical Sciences 31 (2021), 25-56

Sean M. Stewart generalised $$\int_{0}^{\infty}\arctan \left(\frac{2rx}{b^2+r^2}\right)\left(\frac{x}{x^2+a^2}\right)dx=\frac{\pi}{2}\ln\left(\frac{a+\sqrt{r^2+b^2}+r}{a+\sqrt{r^2+b^2}-r}\right)$$

• Thank you for the link to Scienta.
– FDP
Sep 12 at 10:22