# How do you integrate f(x)=(x^3+5x)/((x^2-1)(x-3)(x+9)) using partial fractions?

Apr 11, 2016

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

$= - \frac{3}{20} \ln \left\mid x - 1 \right\mid - \frac{3}{32} \ln \left\mid x + 1 \right\mid + \frac{7}{16} \ln \left\mid x - 3 \right\mid + \frac{129}{160} \ln \left\mid x + 9 \right\mid + C$

#### Explanation:

The denominator factors into distinct linear factors as:

$\left(x - 1\right) \left(x + 1\right) \left(x - 3\right) \left(x + 9\right)$

So we can look for a partial fraction decomposition of the form:

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

$= \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{x - 3} + \frac{D}{x + 9}$

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

$= \frac{A \left({x}^{3} + 7 {x}^{2} - 21 x - 27\right) + B \left({x}^{3} + 5 {x}^{2} - 33 x + 27\right) + C \left({x}^{3} + 9 {x}^{2} - x - 9\right) + D \left({x}^{3} - 3 {x}^{2} - x + 3\right)}{\left({x}^{2} - 1\right) \left(x - 3\right) \left(x + 9\right)}$

$= \frac{\left(A + B + C + D\right) {x}^{3} + \left(7 A + 5 B + 9 C - 3 D\right) {x}^{2} + \left(- 21 A - 33 B - C - D\right) x + \left(- 27 A + 27 B - 9 C + 3 D\right)}{\left({x}^{2} - 1\right) \left(x - 3\right) \left(x + 9\right)}$

Equating coefficients, we get the following system of linear equations:

$\left\{\begin{matrix}A + B + C + D = 1 \\ 7 A + 5 B + 9 C - 3 D = 0 \\ - 21 A - 33 B - C - D = 5 \\ - 27 A + 27 B - 9 C + 3 D = 0\end{matrix}\right.$

This can be expressed in matrix form as:

$\left(\begin{matrix}1 & 1 & 1 & 1 & 1 \\ 7 & 5 & 9 & - 3 & 0 \\ - 21 & - 33 & - 1 & - 1 & 5 \\ - 27 & 27 & - 9 & 3 & 0\end{matrix}\right)$

Perform a sequence of row operations on the matrix to get the left hand $4 \times 4$ submatrix into the form of an identity matrix. Then the right hand column will contain the values of $A , B , C$ and $D$.

Add/subtract multiples of the first row to/from the others to get:

$\left(\begin{matrix}1 & 1 & 1 & 1 & 1 \\ 0 & - 2 & 2 & - 10 & - 7 \\ 0 & - 12 & 20 & 20 & 26 \\ 0 & 54 & 18 & 30 & 27\end{matrix}\right)$

Add/subtract multiples of the second row to/from the third and fourth rows to get:

$\left(\begin{matrix}1 & 1 & 1 & 1 & 1 \\ 0 & - 2 & 2 & - 10 & - 7 \\ 0 & 0 & 8 & 80 & 68 \\ 0 & 0 & 72 & - 240 & - 162\end{matrix}\right)$

Subtract $9$ times the third row from the fourth row to get:

$\left(\begin{matrix}1 & 1 & 1 & 1 & 1 \\ 0 & - 2 & 2 & - 10 & - 7 \\ 0 & 0 & 8 & 80 & 68 \\ 0 & 0 & 0 & - 960 & - 774\end{matrix}\right)$

Divide the $2$nd, $3$rd and $4$th rows by values to make the diagonal all $1$'s:

$\left(\begin{matrix}1 & 1 & 1 & 1 & 1 \\ 0 & 1 & - 1 & 5 & \frac{7}{2} \\ 0 & 0 & 1 & 10 & \frac{17}{2} \\ 0 & 0 & 0 & 1 & \frac{129}{160}\end{matrix}\right)$

Subtract row $2$ from row $1$ to get:

$\left(\begin{matrix}1 & 0 & 2 & - 4 & - \frac{5}{2} \\ 0 & 1 & - 1 & 5 & \frac{7}{2} \\ 0 & 0 & 1 & 10 & \frac{17}{2} \\ 0 & 0 & 0 & 1 & \frac{129}{160}\end{matrix}\right)$

Add/subtract multiples of row $3$ to/from rows $1$ and $2$ to get:

$\left(\begin{matrix}1 & 0 & 0 & - 24 & - \frac{39}{2} \\ 0 & 1 & 0 & 15 & 12 \\ 0 & 0 & 1 & 10 & \frac{17}{2} \\ 0 & 0 & 0 & 1 & \frac{129}{160}\end{matrix}\right)$

Add/subtract multiples of row $4$ to/from rows $1$, $2$ and $3$ to get:

$\left(\begin{matrix}1 & 0 & 0 & 0 & - \frac{3}{20} \\ 0 & 1 & 0 & 0 & - \frac{3}{32} \\ 0 & 0 & 1 & 0 & \frac{7}{16} \\ 0 & 0 & 0 & 1 & \frac{129}{160}\end{matrix}\right)$

Hence:

$\left\{\begin{matrix}A = - \frac{3}{20} \\ B = - \frac{3}{32} \\ C = \frac{7}{16} \\ D = \frac{129}{160}\end{matrix}\right.$

So:

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

$= \int \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{x - 3} + \frac{D}{x + 9} \mathrm{dx}$

$= - \frac{3}{20} \ln \left\mid x - 1 \right\mid - \frac{3}{32} \ln \left\mid x + 1 \right\mid + \frac{7}{16} \ln \left\mid x - 3 \right\mid + \frac{129}{160} \ln \left\mid x + 9 \right\mid + \text{constant}$