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Question Number 194759 by horsebrand11 last updated on 15/Jul/23

$${\int }_0^1{\int }_0^1\left(\frac{1+{\mathrm{x}}^2}{1+{\mathrm{x}}^2+{\mathrm{y}}^2}\right)\mathrm{dxdy}$$

Commented by Frix last updated on 15/Jul/23

$$\mathrm{I}\mathrm{get}\frac{1}{2}+\frac{\sqrt{2}{\tan }^{-1}\sqrt{2}}{4}$$

Answered by dimentri last updated on 15/Jul/23

$${\int }_0^1{\int }_0^1\left(\frac{1+x^2}{1+x^2+y^2}\right)dxdy$$$$=\frac{1}{2}{\int }_0^1{\int }_0^1(1+\frac{1}{1+x^2+y^2})dxdy$$$$=\frac{1}{2}+\frac{1}{2}{\int }_0^1{\int }_0^1\left(\frac{1}{1+x^2+y^2}\right)dxdy$$$$=\frac{1}{2}+{\int }_0^1{\int }_0^x\left(\frac{1}{1+x^2+y^2}\right)dxdy$$$$=\frac{1}{2}+{\int }_0^{\pi /4}{\int }_0^{\sec \theta }\left(\frac{1}{1+r^2}\right)rdrd\theta$$$$=\frac{1}{2}+\frac{1}{2}{\int }_0^{\pi /4}\ln (1+\sec {}^2\theta )d\theta$$$$=\frac{1}{2}+\frac{1}{4}(2G-\pi \ln 2)$$

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