Double Integral $\int_{0}^{4} \int_{\sqrt{x}}^{2} \frac{1}{1+y^3} \mathrm{d}y\;\mathrm{d}x$ I am having trouble computing the double integral:
$$
\int_{0}^{4} \int_{\sqrt{x}}^{2} \frac{1}{1+y^3} \mathrm{d}y\,\mathrm{d}x
$$
I computed the inner integral:
$$
\left [ \frac{1}{3}\ln|y + 1| - \frac{1}{6}\ln(y^2 - y + 1) + \sqrt\frac{1}{3}\tan^{-1}[\frac{(2y-1)}{\sqrt(3)}] \right ]_{\sqrt{x}}^{2}
$$
So now would I just continue as:
$$
\left [ \frac{1}{3}\ln|2 + 1| - \frac{1}{6}\ln(2^2 - 2 + 1) + \sqrt\frac{1}{3}\tan^{-1}[\frac{(2(2)-1)}{\sqrt{3}}] \right ] - \left[\frac{1}{3}\ln|\sqrt{x} + 1| - \frac{1}{6}\ln(\sqrt{x}^2 - \sqrt{x} + 1) + \sqrt\frac{1}{3}\tan^{-1}[\frac{(2\sqrt{x}-1)}{\sqrt 3}]\right]
$$
To finish the inner integral?

If you draw the region of integration, the integral you have is equal to this integral:
$$\int_0^2\int_0^{y^2} \frac{1}{1+y^3} \operatorname{d}x\operatorname{d}y.$$
This is a bit easier to integrate. *After integrating with respect to $x$, you can use a $u$ substitution, $u=1+y^3$.
