[-InnoXenT-]

Let $x = 2\tan{u}$ ---> $dx = 2\sec^2{u}du$ so the integrand will be

$$\int_{0}^{\frac{\pi}{4}} \frac{2\sec^2{u}}{4sec^2{u}} \, du$$

$$= \frac{1}{2}\int_{0}^{\frac{\pi}{4}}\, du$$

$$= \frac{\pi}{8}$$