<s>[INST] 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
$$\begin{aligned}y=\frac{1}{6}\log\left(\frac{x^2-x+1}{(x+1)^2}\right)+\frac{1}{\sqrt3}\arctan\frac{2x-1}{\sqrt3}+C\end{aligned}$$
 [/INST] 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 $\displaystyle x^2-x+1=\frac{(2x-1)^2+(\sqrt3)^2}4$
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