Basic Integral Calculus Evaluate:
$$\int\frac{–4x^3(x^4–2)^{1/3}}{(x^2+2)(x^2–2)}dx$$
I had so many attempt in solving this equation but I can't find the correct answer .
I am asking anybody's help. Please help me.
Thank you!

Expanding from @szw1710's hint
Let $x^4-2 = t^3$ then $4x^3\ dx = 3t^2\ dt$ and so
$$\begin{align}  
\int\frac{(x^4-2)^{1/3}}{x^4 - 4}\ 4x^3dx &= \int \frac{3t^3}{t^3-2}\ dt \\
&= \int \left(3 + \frac{6}{t^3-2}\right)\ dt \\ &= 3t + \int \frac{6}{t^3-2} \ dt
\end{align}$$
Then let $t = \sqrt[3]{2}u$ we get
$$ \int \frac{6}{t^3 -2}\ dt = \int\frac{3\sqrt[3]{2}}{u^3-1}\ du$$
Now use partial fractions to decompose
$$ \frac{1}{u^3-1} = \frac{A}{u-1} + \frac{B(2u+1)}{u^2+u+1} + \frac{C}{\left(u+\frac{1}{2}\right)^2+\frac{3}{4}} $$
which integrates to
$$ \int \frac{1}{u^3-1}\ du = A\ln \big|u-1\big| + B\ln \big|u^2 + u + 1\big| + \frac{2}{\sqrt{3}}C\arctan \left( \frac{2u+1}{\sqrt{3}} \right) $$
The result is, how you say, not pretty... But it's doable.
