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.

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.
