Consider the function on the real numbers $f(x):=\arctan(\frac x{\sqrt{1+x^4}})$. One can easily find that $f’(x)=\frac{1-x^4}{(1+x^2+x^4)\sqrt{1+x^4}}$.
I was wondering how one could compute the anti-derivative of $f’$ without further knowledge about $f$.
That is, how can one compute $$\int \frac{1-x^4}{(1+x^2+x^4)\sqrt{1+x^4}} \, \mathrm dx \text{ ?}$$
A small remark: I chose this function $f’$ as an example of a function that Wolfram Alpha is not able to integrate.
Hint:
Set $x^2=y,2x\ dx= dy$ for $x\ge0$
$$I=\int\dfrac{1-x^4}{(1+x^2+x^4)\sqrt{1+x^4}}dx=\int\dfrac{\dfrac1{x^4}-1}{\left(\dfrac1{x^2}+1+x^2\right)\sqrt{x^2+\dfrac1{x^2}}}x\ dx$$
$$2I=\int\dfrac{\dfrac1{y^2}-1}{\left(y+1+\dfrac1y\right)\sqrt{y+\dfrac1y}}dy$$
Now $\sqrt{y+\dfrac1y}=u\implies du=?, y+\dfrac1y=u^2$
Can you take it from here?