Question: $$\int \frac{x(2-x^3)}{(x^3+1)^2}\,\text{d}x.$$

Is there some simple ways to solve this integral? As my solution for this integral is very long. It's not suitable for my student.

Answer: 
There is a tricky substitution here $$\begin{align}\int \frac{x(2-x^3)}{(x^3+1)^2}dx &=\int \frac{2-x^3}{x^3(\color{blue}{x+\frac{1}{x^2}})^2}dx\end{align}$$

$\color{blue}{x+\frac{1}{x^2}}=u \implies \dfrac{2-x^3}{x^3}dx=-du$ and the integral becomes
$$\begin{align} &-\int \frac{1}{\color{blue}{u^2}}du\end{align}$$
The fastest method seems to rely on knowing what the answer is ahead of time.

If you rewrite the numerator as $x(2-x^3) = 2x(x^3+1)- x^2(3x^2)$, then you see that the integrand $\frac{2x(x^3+1)- x^2(3x^2)}{(x^3+1)^2}$ is precisely the derivative of $\frac{x^2}{x^3+1}$ due to the quotient rule.

This seems not to be good as a general integration technique, but the standard method of partial fractions will require solving for 6 unknowns and will be pretty unwieldy.