Prove $\int_0^1dx\int_x^{1/x}\frac{y^2dy}{(x+y)^2\sqrt{1+y^2}}=\sqrt{2}-\frac{1}{2}$ by change of order of integration. I was trying to prove the following by changing the order of integration:
$$\int_0^1dx\int_x^{1/x}\frac{y^2dy}{(x+y)^2\sqrt{1+y^2}}=\sqrt{2}-\frac{1}{2}$$
Splitting the region $R=\big\{0\leq x\leq1, x\leq y\leq \frac{1}{x}\big\}$ of integration into $R_1=\big\{0\leq x\leq y, 0\leq y\leq1\big\}$ and $R_2=\Big\{1\leq y\leq\infty, 0\leq x\leq\frac{1}{y}\Big\}$ and integrating by changing the order of integration on these two regions, I get the integral to equal $\frac{\sqrt{2}}{2}$.
While using the regions as done in the answer of To prove $\int_0^1dx\int_x^{1/x}\frac{ydy}{(1+xy)^2(1+y^2)}=\frac{\pi-1}{4}$, I get $\frac{\sqrt{2}}{2}+\frac{1}{2}$.
Any suggestion/help would be appreciated. Thanks!
 A: Without seeing your computation, it is not possible to pinpoint your error.  Let $$f(x,y) = \frac{y^2}{(x+y)^2\sqrt{1+y^2}} .$$
Then $$\int f(x,y) \, dx = -\frac{y^2}{(x+y)\sqrt{1+y^2}} + C$$ and in particular, for $g > 0$,
$$F(g,y) = \int_{x=0}^g f(x,y) \, dx = \frac{y^2}{\sqrt{1+y^2}}\left(\frac{1}{y} - \frac{1}{g+y}\right).$$ therefore, the given integral equals
$$\begin{align}
I &= \int_{x=0}^1 \int_{y=x}^{1/x} f(x,y) \, dy \, dx \\
&\int_{y=0}^1 \int_{x=0}^y f(x,y) \, dx \, dy  + \int_{y=1}^\infty \int_{x=0}^{1/y} f(x,y) \, dx \, dy \\
&= \int_{y=0}^1 F(y,y) \, dy + \int_{y=1}^\infty F(1/y,y) \, dy \\
&= \int_{y=0}^1 \frac{y}{2\sqrt{1+y^2}} \, dy + \int_{y=1}^\infty \frac{y}{(1+y^2)^{3/2}} \, dy.
\end{align}$$
The common substitution $$u = 1 + y^2, \quad du = 2y \, dy,$$ easily yields
$$\begin{align}
I &= \frac{1}{4}\int_{u=1}^2 u^{-1/2} \, du + \frac{1}{2} \int_{u=2}^\infty u^{-3/2} \, du \\
&= \frac{1}{4}\left[2u^{1/2} \right]_{u=1}^2 + \frac{1}{2} \left[-2u^{-1/2}\right]_{u=2}^\infty \\
&= \frac{1}{4}\left(2 \sqrt{2} - 2\right) + \frac{1}{2} \left(0 + \frac{2}{\sqrt{2}}\right) \\
&= \frac{\sqrt{2} - 1}{2} + \frac{1}{\sqrt{2}} \\
&= \sqrt{2} - \frac{1}{2}
\end{align}$$
as claimed.
