I'm trying to crack an integral problem whose answer has been lost:
$$ I:=\int_0^\frac{1}{\sqrt{2}}\int_\sqrt{1-x^2}^\sqrt{3-x^2}\frac{x}{1+x^2+y^2}dydx+\int_\frac{1}{\sqrt{2}}^\sqrt{\frac{3}{2}}\int_x^\sqrt{3-x^2}\frac{x}{1+x^2+y^2}dydx$$
Judging from its form, I guess the method of polar coordinates might do the trick. Then I made a drawing to help find the limits of integration:
In this drawing, the left one is responsible for the first integral while the right one helps us take care of the second integral. With this drawing, I transformed the original question into one that reads: $$ I=\int_{\theta=\frac{\pi}{4}}^{\theta=\frac{\pi}{2}}\int_{r=1}^{r=\sqrt{3}}\frac{r\cos\theta}{1+r^2}rdrd\theta+\int_{\theta=\frac{\pi}{4}}^{\theta=\cos^{-1}\frac{1}{\sqrt{6}}}\int_{r=\frac{\cos\theta}{\sqrt{2}}}^{r=\sqrt{\frac{3}{2}}\cos\theta}\frac{r\cos\theta}{1+r^2}rdrd\theta$$
As it looks quite ugly, I have little confidence in what I just obtained. Does anyone get the same result? Thank you.
Thanks to JanG's valuable comment, we can see $$\begin{align} I&=\int_{\theta=\frac{\pi}{4}}^{\theta=\frac{\pi}{2}}\int_{r=1}^{r=\sqrt{3}}\frac{r\cos\theta}{1+r^2}rdrd\theta=\left[\int_{1}^{\sqrt{3}}\left(1-\frac{1}{1+r^2}\right)dr\right]\cdot\left[\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos\theta d\theta\right]\\ &=\left[r-\tan^{-1}r\right]_{1}^{\sqrt{3}}\cdot\left[\sin\theta\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}=\left[(\sqrt 3-1)-(\frac{\pi}{3}-\frac{\pi}{4})\right]\cdot\left(1-\frac{\sqrt 2}{2}\right). \end{align}$$ The computation above suggests my original result will differ from the actual value by an extra integral (the second one).