# How do you integrate int (4x^2 - 7x - 12) / ((x) (x+2) (x-3)) using partial fractions?

Mar 10, 2018

$2 \ln \left\mid x \right\mid + \frac{9}{5} \ln \left\mid x + 2 \right\mid + \frac{1}{5} \ln \left\mid x - 3 \right\mid + C$

#### Explanation:

$\frac{4 {x}^{2} - 7 x - 12}{\left(x\right) \left(x + 2\right) \left(x - 3\right)} = \frac{A}{x} + \frac{B}{x + 2} + \frac{C}{x - 3} =$

$= \frac{A \left(x + 2\right) \left(x - 3\right) + B \left(x\right) \left(x - 3\right) + C \left(x\right) \left(x + 2\right)}{\left(x\right) \left(x + 2\right) \left(x - 3\right)}$
then:

$A \left({x}^{2} - x - 6\right) + B \left({x}^{2} - 3 x\right) + C \left({x}^{2} + 2 x\right) = 4 {x}^{2} - 7 x - 12$

cons. $x = 0$ then $A \left(- 6\right) + B \left(0\right) + C \left(0\right) = - 12 \Rightarrow A = 2$

${x}^{2} - x - 6 = 0 \Rightarrow {x}_{1} = 3 \mathmr{and} {x}_{2} = - 2$

cons. $x = - 2$ then $A \left(0\right) + B \left(10\right) + C \left(0\right) = 18 \Rightarrow B = \frac{9}{5}$

cons. $x = 3$ then $A \left(0\right) + B \left(0\right) + C \left(15\right) = 3 \Rightarrow C = \frac{1}{5}$

finally get to:

$\frac{4 {x}^{2} - 7 x - 12}{\left(x\right) \left(x + 2\right) \left(x - 3\right)} = \frac{2}{x} + \frac{9}{5 \left(x + 2\right)} + \frac{1}{5 \left(x - 3\right)}$

then:

$\int \frac{4 {x}^{2} - 7 x - 12}{\left(x\right) \left(x + 2\right) \left(x - 3\right)} \mathrm{dx} = \int \left(\frac{2}{x} + \frac{9}{5 \left(x + 2\right)} + \frac{1}{5 \left(x - 3\right)}\right) \mathrm{dx} = \int \frac{2}{x} \mathrm{dx} + \int \frac{9}{5 \left(x + 2\right)} \mathrm{dx} + \int \frac{1}{5 \left(x - 3\right)} \mathrm{dx} =$

$2 \ln \left\mid x \right\mid + \frac{9}{5} \ln \left\mid x + 2 \right\mid + \frac{1}{5} \ln \left\mid x - 3 \right\mid + C$