Help me to calculate $\int_{0}^{\pi/4}\sqrt{1+\tan x}\,\mathrm dx$

How to calculate $$\int_{0}^{\pi/4}\sqrt{1+\tan x}\,\mathrm dx$$ My attempt: Let $$I=\int_{0}^{\pi/4}\sqrt{1+\tan x}\,\mathrm dx$$ substitute $\sqrt{1+\tan x}=t$,then $$I=\int_{1}^{\sqrt{2}}\frac{2t^{2}}{t^{4}-2t^{2}+2}\,\mathrm dt=\int_{1}^{\sqrt{2}}\frac{2}{t^{2}-2+\dfrac{2}{t^{2}}}\,\mathrm dt$$ but I got stuck here for a long time.Any idea?

Hint: Try to use partial fractions $$\displaystyle\frac{2t^2}{t^4-2t^2+2}=\frac{At+B}{t^2-\sqrt{2+2\sqrt{2}}t+\sqrt{2}}+\frac{Ct+D}{t^2+\sqrt{2+2\sqrt{2}}t+\sqrt{2}}$$