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

∫_0 ^( 1) ∫_0 ^( 1) (((1+x^2 )/(1+x^2 +y^2 ))) dxdy

$$\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\left(\frac{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\right)\:\mathrm{dxdy}\: \\ $$

Commented by Frix last updated on 15/Jul/23

I get (1/2)+(((√2)tan^(−1) (√2))/4)

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

Answered by dimentri last updated on 15/Jul/23

∫_0 ^( 1) ∫_0 ^( 1) (((1+x^2 )/(1+x^2 +y^2 )))dxdy = (1/2)∫_0 ^1 ∫_0 ^1 (1+(1/(1+x^2 +y^2 )))dxdy = (1/2) +(1/2)∫_0 ^1 ∫_0 ^1 ((1/(1+x^2 +y^2 )))dxdy = (1/2)+∫_0 ^1 ∫_0 ^( x) ((1/(1+x^2 +y^2 )))dxdy = (1/2)+∫_0 ^( π/4) ∫_0 ^( sec θ) ((1/(1+r^2 )))r dr dθ = (1/2)+(1/2)∫_0 ^( π/4) ln (1+sec^2 θ)dθ = (1/2)+(1/4)(2G−π ln 2)

$$\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right){dxdy}\: \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right){dxdy} \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right){dxdy} \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}+\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\:{x}} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right){dxdy}\: \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}+\int_{\mathrm{0}} ^{\:\pi/\mathrm{4}} \int_{\mathrm{0}} ^{\:\mathrm{sec}\:\theta} \left(\frac{\mathrm{1}}{\mathrm{1}+{r}^{\mathrm{2}} }\right){r}\:{dr}\:{d}\theta \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\pi/\mathrm{4}} \mathrm{ln}\:\left(\mathrm{1}+\mathrm{sec}\:^{\mathrm{2}} \theta\right){d}\theta \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{G}−\pi\:\mathrm{ln}\:\mathrm{2}\right)\: \\ $$

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