How do I find this integral $\int\frac{dx}{2x^4+2x^2-1}$

Question

How I evaluate the above integral?

$$\displaystyle\int\dfrac{dx}{2x^4+2x^2-1}$$

I have unsuccessfully tried it more than once. Is there a small substitution that I am missing?

And is there any general approach to problems like this where the denominator is $ax^4+bx^2+c$ where $ax^2+bx+c=0$ does not have any real roots?

Answer 1

Hint

$$\displaystyle\int\dfrac{dx}{2x^4+2x^2-1}=2\displaystyle\int\dfrac{dx}{4x^4+4x^2-2}=2\displaystyle\int\dfrac{dx}{(2x^2+1)^2-3}$$

Answer 2

Hint:It easy cheek that $2x^4+2x^2-1=\frac{1}{2}(2x^2+1+\sqrt{3})(2x^2+1-\sqrt{3})$