# How do you integrate int (x^3 + 2x - 1) / (2x^2 - 3x - 2) using partial fractions?

Mar 7, 2016

${x}^{2} / 4 + \frac{3 x}{4} + \frac{17}{40} \ln | x + \frac{1}{2} | + \frac{11}{5} \ln | x - 2 | + c o n s t .$

#### Explanation:

First we should get in the numerator a polynomial of a grade inferior than the denominator's

By long division:

$\text{ "x^3+0x^2+2x-1" }$|$\text{ } 2 {x}^{2} - 3 x - 2$
$- {x}^{3} + \frac{3}{2} {x}^{2} + x \text{ }$|____
_____" "1/2x+3/4
$\text{ } \frac{3}{2} {x}^{2} + 3 x - 1$
$\text{ } - \frac{3}{2} {x}^{2} + \frac{9}{4} x + \frac{3}{2}$
$\text{ }$ _______
$\text{ } \frac{21}{4} x + \frac{1}{2} = \frac{1}{4} \left(21 x + 2\right)$

So the expression becomes
$= \int \left(\frac{x}{2} + \frac{3}{4}\right) \mathrm{dx} + \frac{1}{4} \int \frac{21 x + 2}{2 {x}^{2} - 3 x - 2} \mathrm{dx}$

Let's deal with the last part
Finding the zeros of the denominator
$2 {x}^{2} - 3 x - 2 = 0$

$\Delta = 9 + 16 = 25$ => $\sqrt{\Delta} = 5$
$x = \frac{3 \pm 5}{4}$ => ${x}_{1} = - \frac{1}{2}$; ${x}_{2} = 2$

Then we can break the second integrand in this way:
$\frac{21 x + 2}{2 {x}^{2} - 3 x - 2} = \frac{A}{x + \frac{1}{2}} + \frac{B}{x - 2}$
To find $A$ and $B$ let's make $x = 0 \mathmr{and} - 1$

$- 1 = 2 A - \frac{B}{2}$
$- \frac{19}{3} = - 2 A - \frac{B}{3}$
Summing these 2 expressions we get
$- \frac{22}{3} = - \frac{5}{6} B$ => $B = \frac{44}{5}$
$\to A = \frac{- 1 + \frac{B}{2}}{2} = \frac{- 1 + \frac{22}{5}}{2}$ => $A = \frac{17}{10}$

So the main expression becomes
$= {x}^{2} / 4 + \frac{3 x}{4} + \left(\frac{1}{4}\right) \left(\frac{17}{10} \int \frac{\mathrm{dx}}{x + \frac{1}{2}} + \frac{44}{5} \int \frac{\mathrm{dx}}{x - 2}\right)$
$= {x}^{2} / 4 + \frac{3 x}{4} + \frac{17}{40} \ln | x + \frac{1}{2} | + \frac{11}{5} \ln | x - 2 | + c o n s t .$