Question

Решите двойной интеграл

Answer 1

$\int_{0}^{2}{\mathrm dx}\int_{x^2\over2}^{x}{x\over x^2+y^2}{\mathrm dy}=\int_{0}^{2}x{\mathrm dx}\int_{x^2\over2}^{x}{1\over x^2+y^2}{\mathrm dy}=\int_{0}^{2}x{1\over x}arctg{y\over x}|_{x^2\over2}^{x}{\mathrm dx}=\int_{0}^{2}\left (arctg1-arctg{x\over2} \right ){\mathrm dx}=arctg(1)x|_{0}^{2}-\int_{0}^{2}arctg{x\over2}{\mathrm dx}=[u=arctg{x\over2},du={2\over4+x^2}dx;dv=dx,v=x]={\pi\over2}-xarctg{x\over2}|_{0}^{2}+2\int_{0}^{2}{x\over 4+x^2}{\mathrm dx}={\pi\over2}-{\pi\over2}+\int_{0}^{2}{1\over 4+x^2}{\mathrm dx^2}=\ln|x^2+4||_{0}^{2}=(\ln8-\ln4)=\ln2$