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

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?

• The polynomial $p$ in the denominator does have roots, by the IVT as $p(0)=-1$ and $p(-1)=3,p(1)=3$, indeed I just showed it has at least two. Notice that $p$ is a quadratic in $x^2$ so you can use the quadratic formula to show $x^2 = \frac{-2\pm\sqrt{12}}{4}$ – James Jul 23 '14 at 13:15
• In a sense there is a substitution: $u=x^2$ is used to find the roots of the denominator with the quadratic formula, and then you take the square root. But in the integration there's no substitution, at least not at the beginning: the beginning of this problem is a straightforward but messy partial fractions problem. – Ian Jul 23 '14 at 13:18
• Here is the general approach. For $a\neq 0:$ $$\int\frac{1}{ax^2+bx+c} dx = \begin{cases} \displaystyle \frac{2}{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}} + C & \text{(for }4ac-b^2>0\mbox{)} \\[12pt] \displaystyle -\frac{2}{\sqrt{b^2-4ac}}\,\mathrm{arctanh}\frac{2ax+b}{\sqrt{b^2-4ac}} + C\qquad\color{red}{\text{or}}\\ \displaystyle\frac{1}{\sqrt{b^2-4ac}}\ln\left|\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right| + C & \text{(for }4ac-b^2<0\mbox{)} \\[12pt] \displaystyle -\frac{2}{2ax+b} + C & \text{(for }4ac-b^2=0\mbox{)} \end{cases}$$ – Tunk-Fey Jul 23 '14 at 17:39

$$\displaystyle\int\dfrac{dx}{2x^4+2x^2-1}=2\displaystyle\int\dfrac{dx}{4x^4+4x^2-2}=2\displaystyle\int\dfrac{dx}{(2x^2+1)^2-3}$$
• $2x^2$, you probably mean. – 5xum Jul 23 '14 at 13:19
Hint:It easy cheek that $2x^4+2x^2-1=\frac{1}{2}(2x^2+1+\sqrt{3})(2x^2+1-\sqrt{3})$