Consider the standard normal distribution with cdf F(x) and pdf f(x). What is $\int_0^\infty F(x)d(f(x))$?
In context this is a required step when finding the expectation of rank statistic $E(X_{(3)}), n=3$. The next step in solution relates the asked integral to $\int_0^\infty \frac{e^{-u^2} }{{2\pi}}du$. How did we get there? What is the intuition of integrating a cdf w.r.t. its pdf in general?
Following the hint by Martín-Blas Pérez Pinilla: $$\int_0^\infty F(x) d(f(x)) = [F(x) f(x)]_{x=0}^\infty - \int_0^\infty (f(x))^2 \, dx.$$
The integral on the right-hand side is $\int_0^\infty \frac{e^{-x^2}}{2\pi} \, dx$. (The "$u$" in your solution is not the "$u$" in the integration by parts.)