Compute integral: $\int_{0}^{1}\frac{\sqrt{x+\sqrt{x^2+1}}}{1+x^2}dx$

Question

How to integrate the following definite integral $$ \int_{0}^{1}\frac{\sqrt{x+\sqrt{x^2+1}}}{1+x^2}dx $$

What I have done is

I substitute $t=\sqrt{x+\sqrt{x^2+1}}$, after simplify I got $$ \int\dfrac{t^2}{t^4+1} dt $$

But the integral of the above expression is bit more lengthy.

Answer 1

$$I=\int\frac{t^2}{1+t^4}dt$$

$$\begin{align}I&=\int\frac{1}{t^2+t^{-2}+2-2}dt=\int\frac{1}{(t+t^{-1})^2-2}dt\\ \\ &=\frac{1}{2\sqrt2}\int\frac{1}{t+t^{-1}-\sqrt2}-\frac{1}{t+t^{-1}+\sqrt2}dt\\ \\ &=\frac{1}{2\sqrt2}\int\frac{t}{t^2-\sqrt2~t+1}dt-\frac{1}{2\sqrt2}\int\frac{t}{t^2+\sqrt2~t+1}dt\end{align}$$

Can you proceed from here?

Answer 2

$$ \begin{aligned} \int \frac{t^2}{1+t^4} d t & =\int \frac{1}{t^2+\frac{1}{t^2}} d t \\ & =\frac{1}{2} \int \frac{\left(1+\frac{1}{t^2}\right)+\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}} d t \\ & =\frac{1}{2} \int \frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^2+2}+\frac{1}{2} \int \frac{d\left(t+\frac{1}{t}\right)}{\left(t+\frac{1}{t}\right)^2-2} \\ & =\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+\frac{1}{4 \sqrt{2}} \ln \left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right|+C \end{aligned} $$