Help in finding the integral function. Can somebody provide a hint in finding the following integral? $$\displaystyle \int \dfrac{1}{(x^3+1)^3} \text{ d}x$$ I thought of using partial fractions but that isn't making any sense.
 A: Remember that $x^3+1 = (x+1)(x^2-x+1)$.  That quadratic polynomial cannot be factored using real numbers, since the discriminant ($b^2-4ac\vphantom{\dfrac\int\int}$) is negative.  So
$(x^3+1)^3 = (x+1)^3 (x^2-x+1)^3$.  That should tell you what the partial fraction decomposition is.
To handle $x^2-x+1$, one (of course) completes the square:
\begin{align}
x^2-x+1 & = \left( x^2 - x + \frac 1 4\right) + \frac 3 4 \\[10pt]
& = \left( x - \frac 1 2 \right)^2 + \frac 3 4 \\[10pt]
& = \frac 3 4 \left( 1 + \left( \frac{2x-1}{\sqrt 3} \right)^2 \right).
\end{align}
Then $\tan\theta = \dfrac{2x-1}{\sqrt 3}$ and $1+\tan^2\theta=\sec^2\theta$ and $\sec^2\theta\,d\theta = \dfrac{2\,dx}{\sqrt 3}$, etc. 
A: This following trick, which is essentially the same as @Daniel Fischer's suggestion, reduces your burden of calculating partial fraction decomposition. First, write
$$ \int \frac{dx}{(1+x^3)^3} = \int \frac{1 + x^3}{(1+x^3)^3} \, dx - \int x \cdot \frac{x^2}{(1+x^3)^3} \, dx. $$
Then by integration by parts, we have
$$ \int x \cdot \frac{x^2}{(1+x^3)^3} \, dx = - \frac{1}{6} \frac{x}{(1+x^3)^2} + \frac{1}{6}\int \frac{dx}{(1+x^3)^2}. $$
Plugging this back to the first identity, we get
$$ \int \frac{dx}{(1+x^3)^3} = \frac{1}{6} \frac{x}{(1+x^3)^2} + \frac{5}{6} \int \frac{dx}{(1+x^3)^2}. $$
Similarly, we have
$$ \int \frac{dx}{(1+x^3)^2} = \frac{1}{3} \frac{x}{(1+x^3)^2} + \frac{2}{3} \int \frac{dx}{1+x^3}. $$
Finally, the last integral can be calculated easily by partial fraction decomposition.
