<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] 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
$$\int\frac{1+x^2}{1-x^2}\frac{1}{\sqrt{1+x^4}}dx
=\int \frac{1+x^2}{1-x^2}\frac{\frac{(1-x^2)^2}{\sqrt2(1+x^2)}dt }{(1-x^2)\sqrt{1+t^2}}\\=\frac1{\sqrt2}\int \frac{1}{\sqrt{1+t^2}}dt
= \frac1{\sqrt2}\sinh^{-1}t= \frac{1}{\sqrt{2}} \sinh^{-1}\frac{\sqrt{2}x}{1-x^2}
$$
</s>