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

Question

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.

Answer 1

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.

Answer 2

\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}