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.

Substituting $x=\sinh t$, we get:
$$I=\int_0^{\sinh^{-1}(1)} \frac{\cosh^2 t}{1+\sinh t}dt$$
This integral can be dealt with using the substitution $t=\log u$:
$$I=\frac{1}{2}\int_1^{1+\sqrt{2}} \frac{(u+1/u)^2 du}{u(2+u-1/u)}=\frac{1}{2}\int_1^{1+\sqrt{2}} \frac{(u^2+1)^2 du}{u^2(2u+u^2-1)}= \\ = \frac{1}{2}\int_1^{1+\sqrt{2}} \frac{(u^4+2u^2+1) du}{u^2((u+1)^2-2)}=$$
$$=\frac{1}{2}\int_2^{2+\sqrt{2}} \frac{((v-1)^4+2(v-1)^2+1) dv}{(v-1)^2(v^2-2)}$$
This is a rational integral and we can use partial fractions to evaluate it:
$$\frac{1}{(v-1)^2(v-\sqrt{2})(v+\sqrt{2})}=\frac{A}{v-1}+\frac{B}{(v-1)^2}+\frac{C}{v-\sqrt{2}}+\frac{D}{v+\sqrt{2}}$$
It's a little ugly, but with elementary algebra I'm sure you can finish.
