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Question Number 26544 by kevansuthar@ gmail.com last updated on 26/Dec/17

∫1/x^2 +y^2 dydx

$$\int\mathrm{1}/{x}^{\mathrm{2}} +{y}^{\mathrm{2}} {dydx} \\ $$

Answered by kaivan.ahmadi last updated on 26/Dec/17

x^2 +y^2 =r^2   ∫∫(1/r^2 )rdrdθ=∫∫(dr/r)dθ=∫lnr dθ=  θlnr

$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} \\ $$$$\int\int\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2}} }\mathrm{rdrd}\theta=\int\int\frac{\mathrm{dr}}{\mathrm{r}}\mathrm{d}\theta=\int\mathrm{lnr}\:\mathrm{d}\theta= \\ $$$$\theta\mathrm{lnr} \\ $$

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