Here it is!

Just take $s^2 = \tan(x) => dx = \frac{2s ds }{ 1 + s^4 }$ and $sqrt( \tan(x) ) = s$ because from 0 to $\pi/2$ the integrand is positive.

Then the integrand turns into $\frac{2s^2 ds }{1+s^4}$ (evaluated from 0 to infinity)

Using partial fractions and taking the limit you'll get the answer: $\pi / sqrt(2)$

Note: $sqrt(u)$ means square root of u