# How do you integrate int (3x-2x^2)/((x+9)(x+7)(x+1))  using partial fractions?

Oct 20, 2017

$- \frac{189}{16} \ln | x + 9 | + \frac{119}{12} \ln | x + 7 | - \frac{5}{48} \ln | x + 1 | + C$

#### Explanation:

The simplest way to do this kind of partial fraction integral is to employ the Heaviside Method of partial fraction decomposition.

Here's the idea: Our goal is to take the expression in the integral and split it into 3 separate fractions, each of which is easier to integrate on their own. This means we need to (in this example, at least) find 3 coefficients $A , B , \mathmr{and} C$ such that the following equation is true for every x:

$\frac{3 x - 2 {x}^{2}}{\left(x + 9\right) \left(x + 7\right) \left(x + 1\right)} = \frac{A}{x + 9} + \frac{B}{x + 7} + \frac{C}{x + 1}$

The Heaviside Method is often used to make this process much simpler. It essentially says you should take one factor of the denominator in the original expression at a time, set it equal to 0 to determine that $x$ value (ie, find the root for that expression), then substitute it into the original fraction expression everywhere except that denominator factor. Whatever you come out with is the coefficient of that factor in the separated fraction. Let's see that in practice:

Find A

The denominator in the A term is $\left(x + 9\right)$. If we set that equal to 0 and solve, we find $x = - 9$. We now go to the original fraction in the integral and we substitue $x = - 9$ in for every term except the $\left(x + 9\right)$ factor in the denominator, and simplify (I'm drawing in that term just so you can see we're ignoring it):

$\frac{3 \left(- 9\right) - 2 {\left(- 9\right)}^{2}}{\textcolor{g r a y}{\cancel{\left(x + 9\right)}} \left(- 9 + 7\right) \left(- 9 + 1\right)} = \frac{- 27 - 162}{\left(- 2\right) \left(- 8\right)} = - \frac{189}{16}$

Thus, $A = - \frac{189}{16}$

Find B

Again, the denominator in the B term is $\left(x + 7\right)$. Setting this equal to 0 and solving leads to $x = - 7$. Going back to the original fraction and substituting:

$\frac{3 \left(- 7\right) - 2 {\left(- 7\right)}^{2}}{\left(- 7 + 9\right) \textcolor{g r a y}{\cancel{\left(x + 7\right)}} \left(- 7 + 1\right)} = \frac{- 21 - 98}{\left(2\right) \left(- 6\right)} = \frac{119}{12}$

Thus, $B = \frac{119}{12}$

Find C

Lastly, the denominator in the C term is $\left(x + 1\right)$. Setting this equal to 0 and solving leads to $x = - 1$. Going back to the original fraction and substituting:

$\frac{3 \left(- 1\right) - 2 {\left(- 1\right)}^{2}}{\left(- 1 + 9\right) \left(- 1 + 7\right) \textcolor{g r a y}{\cancel{\left(x + 1\right)}}} = \frac{- 3 - 2}{\left(8\right) \left(6\right)} = - \frac{5}{48}$

Thus, $C = - \frac{5}{48}$

Replace Integral and Solve

Now we have the expansion of fractions we need to solve the problem:

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

$= \int \left(\frac{A}{x + 9} + \frac{B}{x + 7} + \frac{C}{x + 1}\right) \mathrm{dx}$

$= \int \left(\frac{- \frac{189}{16}}{x + 9} + \frac{\frac{119}{12}}{x + 7} - \frac{\frac{5}{48}}{x + 1}\right) \mathrm{dx}$

$= - \frac{189}{16} \int \frac{1}{x + 9} \mathrm{dx} + \frac{119}{12} \int \frac{1}{x + 7} \mathrm{dx} - \frac{5}{48} \int \frac{1}{x + 1} \mathrm{dx}$

$= - \frac{189}{16} \ln | x + 9 | + \frac{119}{12} \ln | x + 7 | - \frac{5}{48} \ln | x + 1 | + C$