# How do you integrate (x^2-x-8)/((x+1)(x^2+5x+6)) using partial fractions?

Sep 26, 2017

$\int \frac{{x}^{2} - x - 8}{\left(x + 1\right) \left(x + 2\right) \left(x + 3\right)}$

$= - 3 \ln \left\mid x + 1 \right\mid + 2 \ln \left\mid x + 2 \right\mid + 2 \ln \left\mid x + 3 \right\mid + C$

#### Explanation:

Note that:

${x}^{2} + 5 x + 6 = \left(x + 2\right) \left(x + 3\right)$

So:

$\frac{{x}^{2} - x - 8}{\left(x + 1\right) \left(x + 2\right) \left(x + 3\right)} = \frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3}$

We can find $A$, $B$ and $C$ using Oliver Heaviside's cover up method:

$A = \frac{{\left(\textcolor{b l u e}{- 1}\right)}^{2} - \left(\textcolor{b l u e}{- 1}\right) - 8}{\left(\left(\textcolor{b l u e}{- 1}\right) + 2\right) \left(\left(\textcolor{b l u e}{- 1}\right) + 3\right)} = \frac{- 6}{2} = - 3$

$B = \frac{{\left(\textcolor{b l u e}{- 2}\right)}^{2} - \left(\textcolor{b l u e}{- 2}\right) - 8}{\left(\left(\textcolor{b l u e}{- 2}\right) + 1\right) \left(\left(\textcolor{b l u e}{- 2}\right) + 3\right)} = \frac{- 2}{- 1} = 2$

$C = \frac{{\left(\textcolor{b l u e}{- 3}\right)}^{2} - \left(\textcolor{b l u e}{- 3}\right) - 8}{\left(\left(\textcolor{b l u e}{- 3}\right) + 1\right) \left(\left(\textcolor{b l u e}{- 3}\right) + 2\right)} = \frac{4}{2} = 2$

So:

$\int \frac{{x}^{2} - x - 8}{\left(x + 1\right) \left(x + 2\right) \left(x + 3\right)}$

$= \int \frac{- 3}{x + 1} + \frac{2}{x + 2} + \frac{2}{x + 3} \textcolor{w h i t e}{.} \mathrm{dx}$

$= - 3 \ln \left\mid x + 1 \right\mid + 2 \ln \left\mid x + 2 \right\mid + 2 \ln \left\mid x + 3 \right\mid + C$

Sep 26, 2017

$\int \setminus \frac{{x}^{2} - x - 8}{\left(x + 1\right) \left({x}^{2} + 5 x + 6\right)} \setminus \mathrm{dx} = - 3 \ln | x + 1 | + 2 \ln | x + 2 | + 2 \ln | x + 3 | + c$

#### Explanation:

We seek:

$I = \int \setminus \frac{{x}^{2} - x - 8}{\left(x + 1\right) \left({x}^{2} + 5 x + 6\right)} \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus \frac{{x}^{2} - x - 8}{\left(x + 1\right) \left(x + 2\right) \left(x + 3\right)} \setminus \mathrm{dx}$

First let us decompose the fraction into partial fractions, which will be of the form:

$\frac{{x}^{2} - x - 8}{\left(x + 1\right) \left(x + 2\right) \left(x + 3\right)} \equiv \frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3}$
$\text{ } = \frac{A \left(x + 2\right) \left(x + 3\right) + B \left(x + 1\right) \left(x + 3\right) + C \left(x + 2\right) \left(x + 1\right)}{\left(x + 1\right) \left(x + 2\right) \left(x + 3\right)}$

${x}^{2} - x - 8 \equiv A \left(x + 2\right) \left(x + 3\right) + B \left(x + 1\right) \left(x + 3\right) + C \left(x + 2\right) \left(x + 1\right)$

Where $A , B$ are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put $x = - 1 \implies 1 + 1 - 8 = A \left(1\right) \left(2\right) \implies A = - 3$
Put $x = - 2 \implies 4 + 2 - 8 = B \left(- 1\right) \left(1\right) \implies B = 2$
Put $x = - 3 \implies 9 + 3 - 8 = C \left(- 1\right) \left(- 2\right) \implies C = 2$

Thus we have:

$I = \int \setminus - \frac{3}{x + 1} + \frac{2}{x + 2} + \frac{2}{x + 3} \setminus \mathrm{dx}$

Which, we can now integrate to get:

$I = - 3 \ln | x + 1 | + 2 \ln | x + 2 | + 2 \ln | x + 3 | + c$