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

[Unemployed]

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

 

[HallsofIvy]

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.)