Problem in reducing a expression In my previous question, I was able to solve it using Weierstrass substitution as the answers pointed out.
But I was keen to know if my expression can be reduced to the given answer.
Desmos proves that the two integrals are indeed equal, here's the proof.
Now how do I reduce this expression, $$\int_{0}^{\sqrt{2}-1}\frac{2}{(1+t)\bigg(\sqrt{(1+t)+\sqrt{(1+t)^2-1}}+\sqrt{(1+t)-\sqrt{(1+t)^2-1}}\bigg)}dt$$
to this,
$$\int_{0}^{\sqrt{2}-1}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt$$
Any help would be really appreciated. Thank you.
 A: Let's denote $$I_1=\int_{0}^{\sqrt{2}-1}\frac{2}{(1+t)\bigg(\sqrt{(1+t)+\sqrt{(1+t)^2-1}}+\sqrt{(1+t)-\sqrt{(1+t)^2-1}}\bigg)}dt$$ and $$I_2=\int_{0}^{\sqrt{2}-1}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt$$
First let's simplify the denominator in the first integral. Since both terms are positive we are able to do the following trick
$$\sqrt{(1+t)+\sqrt{(1+t)^2-1}}+\sqrt{(1+t)-\sqrt{(1+t)^2-1}}=\left(\left(\sqrt{(1+t)+\sqrt{(1+t)^2-1}}+\sqrt{(1+t)-\sqrt{(1+t)^2-1}}\right)^2\right)^{1/2}=\left((1+t)+\sqrt{(1+t)^2-1}+(1+t)-\sqrt{(1+t)^2-1}+2\sqrt{(1+t)^2-(1+t)^2+1}\right)^{1/2}=\left(2(1+t)+2\right)^{1/2}=\sqrt{2(t+2)}$$
Thus the first integral $I_1$ is simplified to $$I_1=\int_0^{\sqrt{2}-1}\frac{2}{(1+t)\sqrt{2(t+2)}}~dt=\left[
\begin{matrix}
x=\sqrt{t+2} & t=0~\rightarrow x=\sqrt{2}\\
t=x^2-2 & t=\sqrt{2}-1 \rightarrow x=\sqrt{\sqrt{2}+1}\\
dt=2x~dx & 
\end{matrix}
\right]=
$$
$$=\int_\sqrt{2}^\sqrt{\sqrt{2}+1}\frac{2\sqrt{2}x}{(x^2-1)x}~dx=2\sqrt{2}\int_\sqrt{2}^\sqrt{\sqrt{2}+1}\frac{dx}{x^2-1}$$
Now let's transform $I_2$: $$I_2=\int_{0}^{\sqrt{2}-1}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt=
\left[\begin{matrix}
t^2=1-\frac{2}{x^2} & t=0 \rightarrow x=\sqrt{2}\\
2t~dt=\frac{4}{x^3}~dx & t=\sqrt{2}-1 \rightarrow x=\sqrt{\sqrt{2}+1}
\end{matrix}\right]=$$
$$=\int_\sqrt{2}^\sqrt{\sqrt{2}+1}\frac{\frac{8}{x^3}}{\left(2-\frac2{x^2}\right)\sqrt{1-\left(1-\frac2{x^2}\right)}}~dx=2\sqrt{2}\int_\sqrt{2}^\sqrt{\sqrt{2}+1}\frac{dx}{x^2-1}
$$
That's it. If we are just to prove that $I_1=I_2$ everything is done. But it's very simple to compute them.
