<s>[INST] Evaluate $\int_0^1\int_y^\sqrt{y}\frac{y}{\sqrt{x^2+y^2}}\,dx\,dy$ Not sure how to simplify this and start working it out. Any help and step through would be much appreciated thanks!
Ive gotten to $$\left[\frac{1}{2}y(x^2+y^2)^\frac{1}{2}+\frac{1}{3}x^3\right]^\sqrt{y}_y$$
Is this step right? And how do I carry on from here?
Thanks
 [/INST] We have $$\int_0^1\int_y^\sqrt{y}\frac{y}{\sqrt{x^2+y^2}}dxdy=\int_0^1\int_{x^2}^{x}\frac{y}{\sqrt{x^2+y^2}}dydx=\int_0^1\left(\sqrt{x^2+y^2}\right)_{x^2}^{x}dx$$ also $$\int_0^1\left(\sqrt{x^2+y^2}\right)_{x^2}^{x}dx=\frac{\sqrt{2}}{2}-\int_0^1x\sqrt{x^2+1}dx$$ now let $t=x^2+1$. From here you can do
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