The problem is the integral of $$\int {\frac{-8 x}{x^4-a^4}}\, dx$$
I factored out the -8 and divided the x.
I tried to use partial fraction decomposition but it wasn't forming into something I could work with. When I tried to group them I had way too many to match the form of the denominator. I feel like I might be using the wrong integration method but I don't know what else I could use.
(Sorry I redrew my work on my computer so I wouldn't have a crappy picture of my notebook)
My Attempt: Part1 Part2/Answer Which gave me the answer: $$4\left(\frac{a^2}{2}\ln |x^2-a^2|-\frac{a^2}{2}\ln |x^2-a^2|\right)+C$$
Thank you so much!
Noting that
$$ I=\int \frac{-8 x}{x^{4}-a^{4}} d x=-4 \int \frac{d\left(x^{2}\right)}{\left(x^{2}\right)^{2}-\left(a^{2}\right)^{2}} $$
Let $y=x^{2}$ and $b=a^{2}$, then
$$\begin{aligned} I &=-4 \int \frac{d y}{y^{2}-b^{2}}\\&=-4 \int \frac{1}{2 b}\left(\frac{1}{y-b} -\frac{1}{y+b}\right) d y \\ &=-\frac{2}{b}(\ln |y-b|-\ln |y+b|)+C \\ &=-\frac{2}{a^{2}} \ln \left|\frac{x^{2}-a^{2}}{x^{2}+a^{2}}\right|+C \\ &=\frac{2}{a^{2}} \ln \left|\frac{x^{2}+a^{2}}{x^{2}-a^{2}}\right|+C \end{aligned}$$