# Evaluate the integral $\int \frac{x}{a+bx^3}\ dx$

How do I solve integral at this form $$\displaystyle\int \frac{x}{a+bx^3}\ dx$$ ?

• Partial fractions, you always have at least one real root so you can factor the denominator. It is certainly messy. – lulu Jul 25 '15 at 20:32
• Making a substitution $u=\sqrt{a/b\,}\,x$ and using partial fractions looks like it should work since, as @lulu said, you can factor the denominator to be a linear polynomial times a quadratic polynomial. – Clayton Jul 25 '15 at 20:36

partial fraction : in general form ,write it like below and find A,B,C $$\frac{x}{a+bx^3}= \\ \frac{x}{(\sqrt{a}+\sqrt{b}x)((\sqrt{a})^2+(\sqrt{b}x)^2-(\sqrt{a}\sqrt{b}x))}=\\ \space \\ \frac{A}{(\sqrt{a}+\sqrt{b}x)}+\frac{Bx+C}{((\sqrt{a})^2+(\sqrt{b}x)^2-(\sqrt{a}\sqrt{b}x))}$$ then you will have logarithm part + log or arctan part (depends on a,b)
$$a+bx^3 = (\sqrta +x\sqrtb)(\sqrta^2 - x\sqrta\sqrtb + x^2\sqrtb^2)$$
So use partial fractions: $$\frac x {a+bx^3} = \frac C {\sqrta +x\sqrtb} + \frac {Dx+E} {\sqrta^2 - x\sqrta\sqrtb + x^2\sqrtb^2}$$