Help for an Integral $ \int \frac{x^2 + 2x -1}{2x^3 +3x^2 -2} dx $

Question

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

I tried to factor the denominator but I couldn't .

Any help is appreciated

Answer 1

I decided to try an easier example, the reciprocal of a nicer cubic. The roots of $$ x^3 - 3 x + 1 $$ are the real numbers $$ u = 2 \cos \frac{2\pi}{9} \; , \; v = 2 \cos \frac{4\pi}{9} \; , \;w = 2 \cos \frac{8\pi}{9} \; . \; $$ I found the integral using partial fractions with symbols. It worked, but the symbols do not disappear. $$ \int \frac{-1}{x^3 - 3 x + 1} = \frac{\log (x-u)}{(w-u)(u-v)} + \frac{\log (x-v)}{(v-w)(u-v)} + \frac{\log (x-w)}{(v-w)(w-u)} + C \; , $$ or $$ \int \frac{-1}{x^3 - 3 x + 1} \; = \; \frac{\log (x-u)}{3(1-u^2)} + \frac{\log (x-v)}{3(1-v^2)} + \frac{\log (x-w)}{3(1-w^2)} \; \; + \; \; C \; . $$

Answer 2

Let $a,b,c$ to be the roots of the cubic $2x^3 +3x^2 -2=0$ (one real root and two non-real complex conjugate roots). So $$\int \frac{x^2 + 2x -1}{2x^3 +3x^2 -2}\, dx=\frac 12 \int \frac{x^2 + 2x -1}{(x-a)(x-b)(x-c)}\, dx$$ Now, using partial fractions $$\frac{x^2 + 2x -1}{(x-a)(x-b)(x-c)}=\frac{a^2+2 a-1}{(a-b) (a-c) (x-a)}+\frac{b^2+2 b-1}{(b-a) (b-c) (x-b)}+\frac{c^2+2 c-1}{(c-a) (c-b) (x-c)}$$ So, three logarithms, two of them with complex arguments. This is not nice at all and I also suspect a typo somewhere (the $-2$ being more than likely $-1$ which would give $a=b=-1$ and $c=\frac 12$. But since there will be a double root, the given partial fraction decomposition does not apply but $$\frac{x^2 + 2x -1}{(x-a)^2(x-c)}=\frac{a^2+2 a-1}{(a-c) (x-a)^2}+\frac{a^2-2 a c-2 c+1}{(a-c)^2 (x-a)}+\frac{c^2+2 c-1}{(a-c)^2 (x-c)}$$