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Question Number 190419 by horsebrand11 last updated on 02/Apr/23

$$\underset{-1}{\stackrel{1}{\int }}\underset{-\sqrt{1-y^2}}{\stackrel{\sqrt{1-y^2}}{\int }}\ln (x^2+y^2+1)dxdy=?$$

Answered by witcher3 last updated on 04/Apr/23

$$\mathrm{x}=\mathrm{rcos}\left(\mathrm{a}\right)$$$$\mathrm{y}=\mathrm{rsin}\left(\mathrm{a}\right)$$$$-1⩽\mathrm{rsin}\left(\mathrm{a}\right)⩽1,\Rightarrow \mathrm{a}\in [-\frac{\pi }{2},\frac{\pi }{2}]$$$$\mid \mathrm{x}\mid ⩽\sqrt{1-{\mathrm{y}}^2}\iff 0⩽{\mathrm{x}}^2+{\mathrm{y}}^2⩽1\iff 0⩽\mathrm{r}⩽1$$$$-\frac{1}{\mathrm{r}}⩽\sin \left(\mathrm{a}\right)⩽\frac{1}{\mathrm{r}}$$$${\int }_0^1{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\mathrm{rln}(1+{\mathrm{r}}^2)\mathrm{dadr}=\frac{\pi }{2}{\int }_0^1\ln (1+{\mathrm{r}}^2){\mathrm{dr}}^2$$$$=\frac{\pi }{2}{\int }_0^1\ln (1+\mathrm{x})\mathrm{dx}=\frac{\pi }{2}.(2\ln \left(2\right)-1)$$$$$$$$$$

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