How can I integrate $ \int \frac{x^2}{1 + x^3} $? 
$$\int \frac{x^2}{1 + x^3}$$

I found this problem in one of my past papers and didn't understand how the solution to this is $ \frac{1}{3} \ln (1+x^3) $. I would really appreciate it if someone guided me through this. I'm sure this is the answer of the integration because I checked it on WolframAlpha and in the mark scheme, but did not understand how to arrive at the answer. Thanks in advance!
 A: Hint:
$$u=x^3+1\implies du=3x^2\,dx$$
A: The most efficient solution has already been covered by Yagna, however it can never hurt to have another method. The method I will take here is without question an extremely overly complicated method for this question, but is good practice for integrals of rational functions as a general method.
Here you have a rational function of the form:
\begin{equation}
 f(x) = \frac{P_n(x)}{Q_m(x)}
\end{equation}
Were $P_n(x)$ and $Q_m(x)$ are real valued polynomials of orders $n$ and $m$ respectively. The first step is to factor $Q_m(x)$. By the fundamental theorem of algebra we can factor $Q_m(x)$ is a product of constant, linear, and quadratic terms. For your question:
\begin{equation} 
f(x) = \frac{x^2}{x^3 + 1} \rightarrow P_n(x) = P_2(x) = x^2,\quad Q_m(x) = Q_3(x) = x^3 + 1
\end{equation}
Here $Q_3(x)$ is factored quite easily:
\begin{equation}
Q_m(x) = Q_3(x) = x^3 + 1 = \left(x + 1\right)\left(x^2 - x + 1\right)
\end{equation}
Thus, 
\begin{equation}
 f(x) = \frac{x^2}{\left(x + 1\right)\left(x^2 - x + 1\right)}
\end{equation}
We now apply a Partial Fraction Decomposition to yield:
\begin{equation}
f(x) = \frac{x^2}{\left(x + 1\right)\left(x^2 - x + 1\right)} = \frac{1}{3}\cdot \frac{1}{x + 1} -  \frac{1}{3}\cdot \frac{2x - 1}{x^2 - x + 1}
\end{equation} 
And hence, 
\begin{align}
 \int  \frac{x^2}{x^3 + 1}\:dx &= \frac{1}{3}\int \frac{1}{x + 1}\:dx +  \frac{1}{3}\int \frac{2x - 1}{x^2 - x + 1}\:dx \\
&= \frac{1}{3}\ln\left|x + 1\right| +  \frac{1}{3} \cdot \ln\left|x^2 - x + 1 \right| + C
\end{align}
Where $C$ is the constant of integration. 
As before, this is overly complicated for this exact situation, but it is a process that is useful in tackling these types of integrals. 
Edit I had an incorrect partial fraction decomposition which was picked by a commenter. Thank you to that person.
