Integrate ln(cube root (2+y^3)) dydx

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SUMMARY

The discussion focuses on the integration of the function ln(cube root(2+y^3)) dydx, specifically with limits of y from sqrt(x) to 1 and x from 1 to 0. The integral is expressed as -∫(x=1 to 0)∫(y=sqrt(x) to 1) ln(√[3]{2+y^3}) dy dx. By changing the order of integration, it simplifies to -∫(y=0 to 1)∫(x=0 to y^2) ln(√[3]{2+y^3}) dx dy. A substitution of ln(√[3]{2+y^3}) to (1/3)ln(2+y^3) further simplifies the integration process.

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Homework Statement



integrate ln(cube root (2+y^3)) dydx with limits
of y =sqrt (x) to 1
of x from 1 to 0


Homework Equations



integral of ln x= x ln x.

This ends up being very convoluted, haven't even begun to insert limits






The Attempt at a Solution

 
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So this is
\int_{x= 1}^0\int_{y= \sqrt{x}}^1 ln(\sqrt[3]{2+ y^3})dydx= -\int_{x= 1}^0\int_{y= \sqrt{x}}^1 ln(\sqrt[3]{2+ y^3})dydx

If you change the order of integration, that becomes
-\int_{y= 0}^1\int_{x= 0}^{y^2} ln(\sqrt[3]{2+ y^3})dxdy

Integrate with respect to x first and you should see a simple substitution.

(Writing ln(\sqrt[3]{2+ y^3})= (1/3)ln(2+ y^3) will also simplify.)
 

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