Evaluate $\int\limits_0^{1}\frac{\sqrt{1+x^2}}{1+x}dx$ 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.

\begin{align}
\int\limits_0^{1}\frac{\sqrt{1+x^2}}{1+x}dx
= &\int\limits_0^{1}\frac{x-1}{\sqrt{1+x^2}}+\frac2{\sqrt{1+x^2}(1+x)}\ {dx}\\
=&\ \left(\sqrt{1+x^2}-\sinh^{-1}x -\sqrt2\sinh^{-1}\frac{1-x}{1+x} \right)_0^1\\
=& \ \frac{\sinh^{-1}1+1}{\sqrt2+1}
\end{align}
