calculus - Evaluate $intlimits_0^{1}frac{sqrt{1+x^2}}{1+x}dx$

Evaluate:

$I=\int\limits_0^1 \frac{\sqrt{1+x^2}}{1+x}dx$

My try:

Let $x=\tan y$ then $dx=(1+\tan^{2} y)dy$
As for the integration limits: if $x=0$ then $y=0$ and if $x=1$ then $y=\frac{π}{4}$

So:

$I=\int\limits_0^{\frac{π}{4}}\frac{1+\tan^{2} y}{(1+\tan y)\cos y}\,dy$

$I=\int\limits_0^{\frac{π}{4}}\frac{1}{\cos^{3} y+\cos^{2} y\sin y}\,dy$

But I don't know how to continue.