how to integrate
$$\int_{0}^{1} \int_{0}^{\infty} \frac{\ln(1 + x^3) \cdot (\ln(1 + y) + \arctan(y))}{(1 + x^2)(1 + y)^2} \,dx\,dy$$
My attempt
$$\Omega = \int_{0}^{1} \int_{0}^{\infty} \frac{\ln(1 + x^3) \cdot (\ln(1 + y) + \arctan(y))}{(1 + x^2)(1 + y)^2} \,dx\,dy$$
$$*\Omega = \int_{0}^{1} \frac{\ln(1 + y) + \arctan(y)}{(1 + y)^2} \,dy \cdot \int_{0}^{\infty} \frac{\ln(1 + x^3)}{1 + x^2} \,dx = I \cdot J$$
$$*I = \int_{0}^{1} \frac{\ln(1 + y) + \arctan(y)}{(1 + y)^2} \,dy$$
$$u = \ln(1+y) + \arctan(y) \\ v = -\frac{1}{1+y}\,dy \quad \Rightarrow \quad du = \frac{1}{1+y} + \frac{1}{1+y^2} \,dy \\dv = \frac{1}{(1+y)^2} \,dy$$
$$\Rightarrow I = -\frac{1}{1 + y} \left(\ln(1 + y) + \arctan(y)\right) \Big|_{0}^{1} + \int_{0}^{1}\frac{1}{1 + y} \left (\frac{1}{(1 + y)^2} + \frac{1}{(1 + y)^3} \,dy \right)= = \frac{1}{2} - \frac{1}{4} \ln(2)$$
I don't know how to assess $J$ and I'm open to other methods.
