Integral of Euler. Here is an integration problem I found in an old book:
Integrate
$$\int\frac{1+x^2}{1-x^2}\frac{dx}{\sqrt{1+x^4}}$$
The integral is attributed to Euler.
My solution is
let $$y=\frac{1+x^2}{1-x^2}$$ then get
$$\frac{1}{\sqrt{2}} \int \frac{ydy}{\sqrt{y^4-1}}$$
After another substitution $z=y^2$ get
$$\frac{1}{2\sqrt{2}}\int\frac{dz}{\sqrt{z^2-1}}=\frac{1}{2\sqrt{2}}\cosh^{-1}z$$ so my answer is
$$\frac{1}{2\sqrt{2}}\cosh^{-1}\left(\frac{1+x^2}{1-x^2}\right)^2$$
However the answer in the back is
$$\frac{1}{\sqrt{2}} \sinh^{-1}\frac{\sqrt{2}x}{1-x^2}$$
After much work (!!!) I have shown that the two forms are equal to a constant.
My question is: How can one solve the original integral to get the alternative answer directly. What substitution to use ?
 A: $$ I=-\int \dfrac{1}{\sqrt{x^4+1}}\dfrac{x^2+1}{x^2-1} dx=$$
$$=-\int \dfrac{1}{\sqrt{\left(x^2-1\right)^2+2x^2}}\dfrac{x^2+1}{x^2-1}  dx$$
$$=-\int\frac{1}{(x^2-1) \sqrt{1+\frac{2x^2}{\left(x^2-1\right)^2}}}\dfrac{x^2+1}{x^2-1}  dx dx$$
$$=-\int\frac{1}{\sqrt{1+\frac{2x^2}{\left(x^2-1\right)^2}}}\dfrac{x^2+1}{(x^2-1)^2}dx$$
Setting $u:=\sqrt{2} \dfrac{x}{x^2-1},$ as we have
$$\dfrac{du}{dx}=\sqrt{2}\dfrac{(x^2-1)-x(2x)}{(x^2-1)^2}=-\sqrt{2}\dfrac{x^2+1}{(x^2-1)^2}$$
We obtain finaly:
$$I=\dfrac{1}{\sqrt{2}}\int \frac{du}{\sqrt{1+u^2}}$$
Now, remember that $(\sinh^{-1})'(u)=\dfrac{1}{\sqrt{1+u^2}}$.
Remark: The initial inspiration came from this reference found by using (https://approach0.xyz/), a powerful search engine for formulas.
A: Knowing the form of the answer, it is natural to apply the substitution $t= \frac{\sqrt2 x}{1-x^2}$, which leads to
$$1+t^2 = \frac{1+x^4}{(1-x^2)^2},\>\>\>\>\>
dt = \frac{\sqrt2 (1+x^2)}{(1-x^2)^2}dx$$
Then
$$\int\frac{1+x^2}{1-x^2}\frac{1}{\sqrt{1+x^4}}dx
=\int \frac{1+x^2}{1-x^2}\frac{\frac{(1-x^2)^2}{\sqrt2(1+x^2)}dt }{(1-x^2)\sqrt{1+t^2}}\\=\frac1{\sqrt2}\int \frac{1}{\sqrt{1+t^2}}dt
= \frac1{\sqrt2}\sinh^{-1}t= \frac{1}{\sqrt{2}} \sinh^{-1}\frac{\sqrt{2}x}{1-x^2}
$$
A: Another shortcut: Let $I=\int\frac{1+x^2}{1-x^2}\frac{dx}{\sqrt{1+x^4}}=\int\frac{1+\frac 1{x^2}}{\frac 1x-x}\frac{dx}{\sqrt{x^2+\frac 1{x^2}}}$ (By dividing $N^r$ and $D^r$ by $x^2$) 
$I= -\int\frac{1+\frac 1{x^2}}{-x+\frac 1x}\frac{dx}{\sqrt{(-x+\frac 1x)^2+2}}$ 
Now substitute $-x+\frac 1x=t$ so that $(1+\frac 1{x^2})dx=-dt$ and it follows that: 
$I=-\int\frac{dt}{t\sqrt{(t^2+2)}}=\frac 1 {\sqrt 2}\ln|\frac {\sqrt 2} t+\sqrt {\frac {2}{t^2}+1}|=\frac 1{\sqrt 2}\sin h^{-1}(\frac {\sqrt 2}t) +C$ and we are done!
