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Question-165183




Question Number 165183 by mathlove last updated on 27/Jan/22
Commented by Ar Brandon last updated on 27/Jan/22
(x^5 +x^2 y^3 +y^2 x^3 +y^5 )=(x^3 +y^3 )(x^2 +y^2 )  =(x+y)(x^2 −xy+y^2 )(x^2 +y^2 )  ln(...)=  ln(x+y)+ln(x^2 −xy+y^2 )+ln(x^2 +y^2 )  lntegration by part with   { ((u(x)=ln(...))),((v(x)=1)) :}
$$\left({x}^{\mathrm{5}} +{x}^{\mathrm{2}} {y}^{\mathrm{3}} +{y}^{\mathrm{2}} {x}^{\mathrm{3}} +{y}^{\mathrm{5}} \right)=\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$$=\left({x}+{y}\right)\left({x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$$\mathrm{ln}\left(…\right)= \\ $$$$\mathrm{ln}\left({x}+{y}\right)+\mathrm{ln}\left({x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} \right)+\mathrm{ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$$\mathrm{lntegration}\:\mathrm{by}\:\mathrm{part}\:\mathrm{with} \\ $$$$\begin{cases}{\mathrm{u}\left({x}\right)=\mathrm{ln}\left(…\right)}\\{\mathrm{v}\left({x}\right)=\mathrm{1}}\end{cases} \\ $$

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