# How do you integrate (x^3-4x-10)/(x^2-x-6) using partial fractions?

Apr 16, 2016

$\ln | x - 3 | + 2 \ln | x + 2 | + c$

#### Explanation:

The first step is to factor the denominator of the function

$\Rightarrow \frac{{x}^{3} - 4 x - 10}{{x}^{2} - x - 6} = \frac{{x}^{3} - 4 x - 10}{\left(x - 3\right) \left(x + 2\right)}$

since the factors are linear then the coefficients of the partial fractions will be constants , say A and B. Writing the function in terms of it's partial fractions.

$\frac{{x}^{3} - 4 x - 10}{\left(x - 3\right) \left(x + 2\right)} = \frac{A}{x - 3} + \frac{B}{x + 2}$

multiplying through by (x-3)(x+2)

${x}^{3} - 4 x - 10 = A \left(x + 2\right) + B \left(x - 3\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(1\right)$

We now have to find the values of A and B .Note that if x = -2 the term with A will be zero and if x = 3 the term with B will be zero. We can make use of this fact in finding A and B.

let x = -2 in (1) :- 10 = -5B $\Rightarrow \textcolor{b l u e}{\text{ B = 2}}$

let x = 3 in (1) : 5 = 5A $\Rightarrow \textcolor{b l u e}{\text{ A = 1 }}$

$\Rightarrow \frac{{x}^{3} - 4 x - 10}{\left(x - 3\right) \left(x + 2\right)} = \frac{1}{x - 3} + \frac{2}{x + 2}$

Integral becomes : $\int \frac{\mathrm{dx}}{x - 3} + \int \frac{2 \mathrm{dx}}{x + 2}$

$= \ln | x - 3 | + 2 \ln | x + 2 | + c$