# How do you integrate int (1-2x^2)/((x+1)(x+6)(x-7))  using partial fractions?

##### 1 Answer
Mar 28, 2016

$\int \frac{1 - 2 {x}^{2}}{\left(x + 1\right) \left(x + 6\right) \left(x - 7\right)} \mathrm{dx}$

$= \frac{1}{40} \ln \left\mid x + 1 \right\mid - \frac{71}{65} \ln \left\mid x + 6 \right\mid - \frac{97}{104} \ln \left\mid x - 7 \right\mid + c$

#### Explanation:

The denominator has already been factored into distinct linear factors for us, so to construct a partial fraction decomposition we just need to find $A$, $B$ and $C$ to satisfy:

$\frac{1 - 2 {x}^{2}}{\left(x + 1\right) \left(x + 6\right) \left(x - 7\right)} = \frac{A}{x + 1} + \frac{B}{x + 6} + \frac{C}{x - 7}$

$= \frac{A \left(x + 6\right) \left(x - 7\right) + B \left(x + 1\right) \left(x - 7\right) + C \left(x + 1\right) \left(x + 6\right)}{\left(x + 1\right) \left(x + 6\right) \left(x - 7\right)}$

$= \frac{A \left({x}^{2} - x - 42\right) + B \left({x}^{2} - 6 x - 7\right) + C \left({x}^{2} + 7 x + 6\right)}{\left(x + 1\right) \left(x + 6\right) \left(x - 7\right)}$

$= \frac{\left(A + B + C\right) {x}^{2} + \left(- A - 6 B + 7 C\right) x + \left(- 42 A - 7 B + 6 C\right)}{\left(x + 1\right) \left(x + 6\right) \left(x - 7\right)}$

Equating coefficients we get the system of linear equations:

$\left\{\begin{matrix}A + B + C = - 2 \\ - A - 6 B + 7 C = 0 \\ - 42 A - 7 B + 6 C = 1\end{matrix}\right.$

Write this out as a matrix:

$\left(\begin{matrix}1 & 1 & 1 & - 2 \\ - 1 & - 6 & 7 & 0 \\ - 42 & - 7 & 6 & 1\end{matrix}\right)$

Perform a sequence of row operations to make the left hand $3 \times 3$ square into the identity matrix. Then the rightmost colum will give us the values of $A$, $B$ and $C$.

Add row $1$ to row $2$ to get:

$\left(\begin{matrix}1 & 1 & 1 & - 2 \\ 0 & - 5 & 8 & - 2 \\ - 42 & - 7 & 6 & 1\end{matrix}\right)$

Add $42 \times$ row $1$ to row $3$ to get:

$\left(\begin{matrix}1 & 1 & 1 & - 2 \\ 0 & - 5 & 8 & - 2 \\ 0 & 35 & 48 & - 83\end{matrix}\right)$

Add $7 \times$ row $2$ to row $3$ to get:

$\left(\begin{matrix}1 & 1 & 1 & - 2 \\ 0 & - 5 & 8 & - 2 \\ 0 & 0 & 104 & - 97\end{matrix}\right)$

Multiply row $2$ by $- \frac{1}{5}$ to get:

$\left(\begin{matrix}1 & 1 & 1 & - 2 \\ 0 & 1 & - \frac{8}{5} & \frac{2}{5} \\ 0 & 0 & 104 & - 97\end{matrix}\right)$

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

$\left(\begin{matrix}1 & 0 & \frac{13}{5} & - \frac{12}{5} \\ 0 & 1 & - \frac{8}{5} & \frac{2}{5} \\ 0 & 0 & 104 & - 97\end{matrix}\right)$

Divide row $3$ by $104$ to get:

$\left(\begin{matrix}1 & 0 & \frac{13}{5} & - \frac{12}{5} \\ 0 & 1 & - \frac{8}{5} & \frac{2}{5} \\ 0 & 0 & 1 & - \frac{97}{104}\end{matrix}\right)$

Subtract $\frac{13}{5} \times$ row $3$ from row $1$ to get:

$\left(\begin{matrix}1 & 0 & 0 & \frac{1}{40} \\ 0 & 1 & - \frac{8}{5} & \frac{2}{5} \\ 0 & 0 & 1 & - \frac{97}{104}\end{matrix}\right)$

Add $\frac{8}{5} \times$ row $3$ to row $2$ to get:

$\left(\begin{matrix}1 & 0 & 0 & \frac{1}{40} \\ 0 & 1 & 0 & - \frac{71}{65} \\ 0 & 0 & 1 & - \frac{97}{104}\end{matrix}\right)$

So:

$\left\{\begin{matrix}A = \frac{1}{40} \\ B = - \frac{71}{65} \\ C = - \frac{97}{104}\end{matrix}\right.$

Then using $\int \frac{1}{t} \mathrm{dt} = \ln \left\mid t \right\mid + C$ we have:

$\int \frac{1 - 2 {x}^{2}}{\left(x + 1\right) \left(x + 6\right) \left(x - 7\right)} \mathrm{dx}$

$= \int \frac{A}{x + 1} + \frac{B}{x + 6} + \frac{C}{x - 7} \mathrm{dx}$

$= \frac{1}{40} \ln \left\mid x + 1 \right\mid - \frac{71}{65} \ln \left\mid x + 6 \right\mid - \frac{97}{104} \ln \left\mid x - 7 \right\mid + c$