# What is int (-x^3+9x-1 ) / (-2x^2- 3 x +5 )?

Dec 8, 2016

The answer is =x^2/4-3/4x-9/8ln(∣2x+5∣)-ln(∣x-1∣)+C

#### Explanation:

Let's do a long division

$\textcolor{w h i t e}{a a a a}$$- {x}^{3} +$$\textcolor{w h i t e}{a a a a a a a a a}$$9 x - 1$$\textcolor{w h i t e}{a a}$∣$- 2 {x}^{2} - 3 x + 5$

$\textcolor{w h i t e}{a a a a}$$- {x}^{3} +$$\textcolor{w h i t e}{a}$$- \frac{3}{2} {x}^{2} + \frac{5}{2} x$$\textcolor{w h i t e}{a a a a a}$∣$\frac{x}{2} - \frac{3}{4}$

$\textcolor{w h i t e}{a a a a a a a}$$0 +$$\textcolor{w h i t e}{a}$$+ \frac{3}{2} {x}^{2} + \frac{13}{2} x - 1$

$\textcolor{w h i t e}{a a a a a a a a a a a}$color(white)(aaa)3/2x^2+9/4x-15/4

$\textcolor{w h i t e}{a a a a a a a a a a a}$color(white)(aaaaa)0+17/4x+11/4

$- 2 {x}^{2} - 3 x + 5 = - \left(2 x + 5\right) \left(x - 1\right)$

So,

$\frac{- {x}^{3} + 9 x - 1}{- {2}^{2} - 3 x + 5} = \left(\frac{x}{2} - \frac{3}{4}\right) + \frac{\frac{17}{4} x + \frac{11}{4}}{- 2 {x}^{2} - 3 x + 5}$

$= \left(\frac{x}{2} - \frac{3}{4}\right) - \frac{\frac{17}{4} x + \frac{11}{4}}{\left(2 x + 5\right) \left(x - 1\right)}$

Let's do a partial fraction decomposition

$\frac{\frac{17}{4} x + \frac{11}{4}}{\left(2 x + 5\right) \left(x - 1\right)} = \frac{A}{2 x + 5} + \frac{B}{x - 1}$

$= \frac{A \left(x - 1\right) + B \left(2 x + 5\right)}{\left(2 x + 5\right) \left(x - 1\right)}$

$\frac{17}{4} x + \frac{11}{4} = A \left(x - 1\right) + B \left(2 x + 5\right)$

Let $x = 1$, $\implies$, $\frac{28}{4} = 7 B$, $\implies$, $B = 1$

Let $x = 0$,$\implies$,$\frac{11}{4} = - A + 5 B$

$A = 5 - \frac{11}{4} = \frac{9}{4}$

So,
$\int \frac{\left(- {x}^{3} + 9 x - 1\right) \mathrm{dx}}{- {2}^{2} - 3 x + 5}$

$= \int \left(\frac{x}{2} - \frac{3}{4}\right) \mathrm{dx} - \frac{9}{4} \int \frac{\mathrm{dx}}{2 x + 5} - \int \frac{\mathrm{dx}}{x - 1}$

=x^2/4-3/4x-9/8ln(∣2x+5∣)-ln(∣x-1∣)+C