Define $f: [0,\infty[\times [0,\infty[ \to \bar{\mathbb R}$, $f(x,y):= xe^{-x^{2}(1+y^{2})}$ it is clear that $f(x,y)\geq 0$ $\lambda-$a.e. Therefore, Tonelli's theorem holds. This means:
$\int_{[0,\infty[} (\int_{[0,\infty[}xe^{-x^{2}(1+y^{2})}d\lambda (x)) d\lambda (y)=\int_{[0,\infty[} (\int_{[0,\infty[}xe^{-x^{2}(1+y^{2})}d\lambda (y)) d\lambda (x)$
Looking exclusively at $\int_{[0,\infty[}xe^{-x^{2}(1+y^{2})}d\lambda (y)$, the only thing that comes to mind is "Riemann Integration", namely:
$\int_{[0,\infty[}xe^{-x^{2}(1+y^{2})}d\lambda (y)=\lim_{n\to \infty}\int_{[0,n]}xe^{-x^{2}(1+y^{2})}dy=\lim_{n\to \infty}\frac{1}{-x\times2y}e^{-x^2(1+y^2)}\vert_{0}^{n}$
Which gets me stuck as I have a $y$ in the denominator, however, I see no other way of doing it. Any other ideas?
Since the Lebesgue integral agrees with the Riemann integral if we have absolutely Riemann-integrable functions we can first integrate with respect to $x$ to get:
$$\int_{0}^{\infty}xe^{-x^2(1+y^2)}d\lambda(x) = \frac{-1}{2}\frac{e^{-x^2(1+y^2)}}{1+y^2}\Big\vert_{0}^{\infty} = \frac{1}{2}\frac{1}{1+y^2}.$$
Now integrate
$$\frac{1}{2}\int_{0}^{\infty}\frac{1}{1+y^2}d\lambda(y)$$
NOTE: A primitive for $xe^{-x^2(1+y^2)}$ with respect to $y$ is not given by $\frac{1}{-x^2\cdot 2y}e^{-xy^2}$! Rather
$$\frac{1}{x}\frac{d}{dy}\frac{1}{2y}e^{-x^2y^2} = \frac{1}{x}\cdot\frac{-2yx^2e^{-x^2y^2}(2y)-2e^{-x^2y^2}}{4y^2}=-xe^{-x^2y^2}-\frac{e^{-x^2y^2}}{2xy^2}$$