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

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.

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    $\begingroup$ Show your solution $\endgroup$ – Empty Apr 6 '15 at 7:19
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    $\begingroup$ The usual way involves partial fraction decomposition, though it's not the fastest here. However, a tricky solution is not really useful to a student. $\endgroup$ – Jean-Claude Arbaut Apr 6 '15 at 7:53
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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 $$\begin{align} &-\int \frac{1}{\color{blue}{u^2}}du\end{align}$$

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    $\begingroup$ Your substitution is not bijective. $\endgroup$ – Jean-Claude Arbaut Apr 6 '15 at 7:50
  • $\begingroup$ Ahh yes, nice catch :) this should fail $\endgroup$ – ganeshie8 Apr 6 '15 at 7:53
  • $\begingroup$ @Jean-ClaudeArbaut maybe we can restrict it to some domain so that it is bijective. $\endgroup$ – Vim Apr 6 '15 at 10:24
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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.

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  • $\begingroup$ You don't need to know the answer ahead of time, but you do need to test out a possibility - the $(x^3+1)^2$ on the denominator and the $x^3$ on the numerator should immediately make you think of the quotient rule. From there, it's a matter of testing to see if the numerator can be expressed appropriately. $\endgroup$ – Glen O Apr 6 '15 at 10:57

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