3

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

Hamid Mohammad
  • 161
  • 1
  • 1
  • 7
  • 4
    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
  • 2
    Making a substitution $u=\sqrt[3]{a/b,},x$ 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

2 Answers2

5

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

Khosrotash
  • 24,922
3

$$ a+bx^3 = (\sqrt[3]a +x\sqrt[3]b)(\sqrt[3]a^2 - x\sqrt[3]a\sqrt[3]b + x^2\sqrt[3]b^2) $$

So use partial fractions: $$ \frac x {a+bx^3} = \frac C {\sqrt[3]a +x\sqrt[3]b} + \frac {Dx+E} {\sqrt[3]a^2 - x\sqrt[3]a\sqrt[3]b + x^2\sqrt[3]b^2} $$

Michael Hardy
  • 1