<s>[INST] 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 ?
 [/INST] 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!
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