Solve ODE $(1+x^3)y'=x$ I was doing problems from Simmons and got stuck at this:
$$(1+x^3)y'=x$$
If it was $x^2$ instead of $x$ then we could simply substitute $1+x^3$ for some variable, but that is not the case. I also tried using
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$
The given answer is
$$\begin{aligned}y=\frac{1}{6}\log\left(\frac{x^2-x+1}{(x+1)^2}\right)+\frac{1}{\sqrt3}\arctan\frac{2x-1}{\sqrt3}+C\end{aligned}$$
 A: We have $$dy=\frac x{1+x^3}dx$$
Using Partial Fraction Decomposition
 $$\frac x{1+x^3}=\frac A{1+x}+\frac{Bx+C}{1-x+x^2}$$
For the ease of calculation, we can write $\displaystyle Bx+C=\frac B2\frac{d(x^2-x+1)}{dx}+D$ so that
$$x=A(1-x+x^2)+(1+x)\left[\frac B2(2x-1)+D\right]$$
Compare the constants, the coefficients of $x,x^2$ to find $A,B,D$ 
Finally use Trigonometric substitution for $\displaystyle x^2-x+1=\frac{(2x-1)^2+(\sqrt3)^2}4$
A: Here's another approach for computing your integral:
First, let's denote $Q(x) = \dfrac{x}{1+x^3}$ your integrand and let $r_i$ be the three roots of the denominator. Clearly, we can see that the equation $1+x^3 = 0$ has $r = \{ -1, e^{\mathrm{i} \, \pi/3}, e^{- \mathrm{i} \,\pi/3}\}$ as solutions, $\mathrm{i}^2 = -1$. Therefore, we can write:
$$Q(x) = \sum_i \frac{A_i}{x-r_i}, \quad A_i \in \mathbb{C}. \tag{1}$$ 
Then, we can compute the $A_i$ terms of the partial fraction decomposition as follows:
$$A_i = \lim_{x \to r_i} (x-r_i) Q(x), \tag{2} $$ where the values of the $A_i$ are given by:

$$A_1 = -1/3, \quad  A_2 = \frac{1}{6} - \frac{\mathrm{i}}{2\sqrt{3}}, \quad A_3 = \overline{A}_2 =   \frac{1}{6} + \frac{\mathrm{i}}{2\sqrt{3}}. \tag{3}$$

We can therefore write your integral as:
$$I = \int Q(x) \, \mathrm{d}x = \sum_i \int\frac{A_i}{x-r_i}\, \mathrm{d}x = \sum_i \text{Log}{|x-r_i|}. \tag{4}$$
The result of the integral must be real, since $Q(x)$ is a real-valued function but we have come up with things like $\log{z}$ where $z$ is a complex number. What can we do now? If we remember the definition of complex logarithm, we have that $\text{Log}{z} = \log{|z|} + \mathrm{i} \, \text{Arg}{z}$ and hence:

\begin{align}
\color{blue}{I} = & \small{- \frac{1}{3} \log{|x+1|} + \left(  \frac{1}{6} - \frac{\mathrm{i}}{2 \sqrt{3}}   \right) \log{x-\frac{1}{2} - \frac{\mathrm{i} \sqrt{3}}{2} } + \left(  \frac{1}{6} + \frac{\mathrm{i}}{2 \sqrt{3}}   \right)  \log{x-\frac{1}{2} - \frac{\mathrm{i} \sqrt{3}}{2} }}\\
 = & -\tiny{ \frac{1}{3} \log{|x+1|} + A_2 \left[ \log{ \sqrt{\left(x-\frac{1}{2}\right)^2 +\frac{3}{4}} + \mathrm{i} \arctan{\left( - \frac{\sqrt{3}}{2x-1} \right)}  } \right]  +  A_3 \left[ \log{ \sqrt{\left(x-\frac{1}{2}\right)^2 +\frac{3}{4}} + \mathrm{i} \arctan{\left( \frac{\sqrt{3}}{2x-1} \right)}  } \right]  } \\
 = & \small{- \frac{1}{3} \log{|x+1|} + \frac{2}{6} \log{\sqrt{x^2 -x +1} } + \frac{1}{2\sqrt{3}} \arctan\left( - \frac{\sqrt{3}}{2x-1} \right)  - \frac{1}{2\sqrt{3}} \arctan\left(  \frac{\sqrt{3}}{2x-1} \right) } \\
  = & \color{blue}{ \frac{1}{6} \log{\frac{x^2-x+1}{(x+1)^2}} + \frac{1}{\sqrt{3}} \arctan{\frac{2x-1}{\sqrt{3}} }} , \qquad \qquad \qquad  \qquad \qquad \qquad \qquad (5)
\end{align}  

where in the last step I have used the fact that $\arctan{1/x} = \pi/2 - \arctan{x}$ and I have take advantage of the oddness character of $\arctan{\theta}$. Of course, never forget about the integration constants!
I hope this helps.
Cheers!
