I post here cause I've got a problem when I'm calculating : $\;I = \int_{D} \frac{1}{1+x^2+y^2} \, \mathrm{d}\lambda(x,y)$ on $D=\bigl\{ \, (x,y) \in \mathbb{R}^2 \mid 0 \leq x \leq 1, 0 \leq y \leq 1, 0 \leq x^2+y^2 \leq 1 \, \bigr\}$.
I know that we can make a change of variables and use the polar coordinates, but I wanted to do it by an other way. So I wanted to make this change of variables : $\phi : (x,y) \rightarrow (x^2+y^2,y)$ ($D \rightarrow \phi(D)$), $\phi$ is injective on $D$ and $\det(\operatorname{Jac}_{\phi}(x,y)) = 2x$ which is different from $0$ if $x$ is different from $0$. So, $\lambda(\{0\} \times \mathbb {R}) = 0$, we can suppose $x \neq 0$. And so, $\phi$ is a diffeomorphism from $D$ (but new $D = \{\, (x,y) \in \mathbb{R}^2 \mid 0 < x \leq 1,\: 0 \leq y \leq 1,\: 0 \leq x^2+y^2 \leq 1 \, \}$) to $\phi(D)$. So i can use the theorem for change of variables, and I have: $$\int_{D} \frac{1}{1+x^2+y^2} \, \mathrm{d}\lambda(x,y)=\int_{\phi(D)} \frac{1}{2x} \times \frac{1}{1+x} \, \mathrm{d}\lambda(x,y)$$
Then, I have to determine $\phi(D)$, and I find that $$\phi(D) = \bigl\{ (u,v) \in \mathbb{R}^2 \mid 0 < u \leq 1 , 0 \leq v \leq 1, v^2 \leq u \bigr\}.$$ So, I use the Fubini-Tonelli's theorem to find : $$I = \frac{1}{2} \int_{0}^{1} \frac{1}{x^2+x} \int_{0}^{\sqrt x}\! \mathrm{d}y\,\mathrm{d}x = \frac{1}{2} \int_{0}^{1}\! \frac{\sqrt x}{x^2+x}\, \mathrm{d}x = \frac{\pi}{4}. $$
But I made a mistake, cause the right result is : $\;I = \log(2)\,\smash[b]{\dfrac{\pi}{4}}$. But I don't know where I made a mistake.
Someone could help me, please ?
Thank you ! (and sorry for my english, I made my best but I've still have some difficulties sometimes)
Hint:
Use polar coordinates: $$\iint_D\frac1{1+x^2+y^2}\,\mathrm d\lambda(x,y)=\int_0^{\tfrac\pi2}\!\!\!\int_0^1\!\!\frac1{1+r^2}\, r\,\mathrm dr\,\mathrm d\theta.$$