How to solve $\int \:\frac{3x^2+1}{x^3+2}dx$?

Question

I am trying to solve the following integral:

$$\int \:\frac{3x^2+1}{x^3+2}dx$$

I am not sure how to do it. I tried using partial fractions but you can not factor $x^3+2$, and changing the variable with $u$ does not seem to work either. Anyone has any ideas?

Answer 1

Hint: write your integral in the form

$$\int \frac{3x^2+1}{(x+\sqrt[3]{2})(x^2-\sqrt[3]{2}x+2^{2/3})}dx$$

Answer 2

Substitute $u=x^3$ and you can reduce it down to:

$$\frac13\int\frac{u^{-\frac23}}{u+2}du+\int\frac{1}{u+2}du$$

I believe it can be taken from here

Answer 3

Wolfy gives a number of forms, the first of which is

$ \log(x^3 + 2) - \dfrac{\log(2^{1/3} x^2 - 2^{2/3} x + 2)}{6\ 2^{2/3}} + \dfrac{\log(2^{2/3} x + 2)}{3\ 2^{2/3}} + \dfrac{\tan^{-1}(\frac{2^{2/3} x - 1}{\sqrt{3}})}{2^{2/3} \sqrt{3}} +C $

Answer 4

$$I=\int \frac{3x^2+1}{x^3+2}dx=\int \frac{3x^2}{x^3+2}dx+\int \frac{dx}{x^3+2}$$ $$I=\log(x^3+2)+\int \frac{dx}{x^3+2}$$ For the remaining integral, let $$x=\sqrt[3]2\, t\implies dx=\sqrt[3]2\,dt$$ $$J=\int \frac{dx}{x^3+2}=\frac{\sqrt[3]2}2 \int \frac{dt}{t^3+1}$$ Now $$t^3+1=(t-1)(t-a)(t-b)$$ Partial fraction decomposition $$\frac 1 {t^3+1}=\frac{1}{(a-1) (a-b) (t-a)}+\frac{1}{(b-1) (b-a) (t-b)}+\frac{1}{(a-1) (b-1) (t-1)}$$ Integrate to get a weighted sum of logarithms with complex values (the weights are complex too) and recombine them to get the answer already given.