I am trying to solve the integral
$$\iint_E\frac{1 + x}{1 + 2x^2 + 3y^2} dA$$
where E is the elliptical area given by $$2x^2 + 3y^2\leq 6$$
I have tried substitution with $(x, y)\mapsto(\sqrt{3} r\cos\theta, \sqrt{2}r\sin\theta)$, where the new ranges are $r\in [0, 1]$ and $\theta\in [0, 2\pi]$. Then I substitute that into the integral and solve it with the new limits.
Is this the right way to go? I get an answer of $\pi\ln 7$ and that seems a bit weird but I am pretty unsure. Can someone guide me in the right direction in this problem? Thanks!
Due to symmetry, the integral simplifies
\begin{align} I= \iint_{E}\frac{1 + x}{1 + 2x^2 + 3y^2} dxdy = \iint_{E}\frac{1}{1 + 2x^2 + 3y^2} dx d y \end{align} Then, with $x=\sqrt3 u$ and $y=\sqrt2 v$ \begin{align} I &= \iint_{2x^2 + 3y^2\leq 6}\frac{1}{1 + 2x^2 + 3y^2} dx d y\\ & =\sqrt6 \iint_{u^2+v^2\leq 1}\frac{1}{1 + 6(u^2 + v^2)} du d v\\ & =2\pi \sqrt6 \int_0^1\frac{r}{1 + 6r^2} dr = \frac{\pi \ln7}{\sqrt6}\\ \end{align}