# How do you integrate int (x-9)/((x+3)(x-7)(x+4))  using partial fractions?

Mar 12, 2016

$\int \left(\frac{x - 9}{\left(x + 3\right) \left(x - 7\right) \left(x + 4\right)}\right) \mathrm{dx}$

= $\frac{132}{110} \ln \left(x + 3\right) - \frac{2}{110} \ln \left(x - 7\right) - \frac{13}{11} \ln \left(x + 4\right) + c$

#### Explanation:

To integrate, we should first convert $\frac{x - 9}{\left(x + 3\right) \left(x - 7\right) \left(x + 4\right)}$ in to partial fractions. Let

$\frac{x - 9}{\left(x + 3\right) \left(x - 7\right) \left(x + 4\right)} \Leftrightarrow \frac{A}{x + 3} + \frac{B}{x - 7} + \frac{C}{x + 4}$. Simplifying RHS

=$\frac{A \left(x - 7\right) \left(x + 4\right) + B \left(x + 3\right) \left(x + 4\right) + C \left(x + 3\right) \left(x - 7\right)}{\left(x + 3\right) \left(x - 7\right) \left(x + 4\right)}$ or

=$\frac{A \left({x}^{2} - 3 x - 28\right) + B \left({x}^{2} + 7 x + 12\right) + C \left({x}^{2} - 4 x - 21\right)}{\left(x + 3\right) \left(x - 7\right) \left(x + 4\right)}$

=$\frac{\left(A + B + C\right) {x}^{2} + \left(- 3 A + 7 B - 4 C\right) x + \left(- 28 A + 12 B - 21 C\right)}{\left(x + 3\right) \left(x - 7\right) \left(x + 4\right)}$

Hence $A + B + C = 0$, $- 3 A + 7 B - 4 C = 1$ and $- 28 A + 12 B - 21 C = - 9$

Solving these will give us $A = \frac{132}{110}$, $B = - \frac{2}{110}$ and $C = - \frac{13}{11}$

Hence $\frac{x - 9}{\left(x + 3\right) \left(x - 7\right) \left(x + 4\right)} \Leftrightarrow \frac{132}{110 \left(x + 3\right)} - \frac{2}{110 \left(x - 7\right)} - \frac{13}{11 \left(x + 4\right)}$

Hence $\int \frac{x - 9}{\left(x + 3\right) \left(x - 7\right) \left(x + 4\right)} \mathrm{dx}$ =

$\int \left[\frac{132}{110 \left(x + 3\right)} - \frac{2}{110 \left(x - 7\right)} - \frac{13}{11 \left(x + 4\right)}\right] \mathrm{dx}$

or

Now one can use the identity $\int \left(\frac{1}{a x + b}\right) \mathrm{dx} = \frac{1}{a} \ln \left(a x + b\right)$

Hence, $\int \left[\frac{132}{110 \left(x + 3\right)} - \frac{2}{110 \left(x - 7\right)} - \frac{13}{11 \left(x + 4\right)}\right] \mathrm{dx}$

= $\frac{132}{110} \ln \left(x + 3\right) - \frac{2}{110} \ln \left(x - 7\right) - \frac{13}{11} \ln \left(x + 4\right) + c$