# How do you integrate f(x)=x/((x-2)(x+4)(x-7)) using partial fractions?

Jan 10, 2016

$I = \frac{1}{13} \ln | x - 2 | - \frac{18}{143} \ln | x + 4 | + \frac{7}{143} \ln | x - 7 | + c$

#### Explanation:

Let's say that there are constant $a$, $b$ and $d$ so

$\frac{a}{x - 2} + \frac{b}{x + 4} + \frac{d}{x - 7} = \frac{x}{\left(x - 2\right) \left(x + 4\right) \left(x - 7\right)}$

Solving it back we have

$\frac{a \left(x + 4\right) \left(x - 7\right) + b \left(x - 2\right) \left(x - 7\right) + d \left(x - 2\right) \left(x + 4\right)}{\left(x - 2\right) \left(x + 4\right) \left(x - 7\right)} = \frac{x}{\left(x - 2\right) \left(x + 4\right) \left(x - 7\right)}$

That also means,

$a {x}^{2} + 4 a x - 7 a x + 28 a + b {x}^{2} - 9 b x + 14 b + {\mathrm{dx}}^{2} + 2 \mathrm{dx} - 8 d = x$

Or

${x}^{2} \left(a + b + d\right) + x \left(4 a - 7 a - 9 b + 2 d\right) + \left(28 a + 14 b - 8 d\right) = x$

Or, still

$a + b + d = 0$
$- 3 a - 9 b + 2 d = 1$
$14 a + 7 b - 4 d = 0$

From there, it's just a question of solving the system. Using Cramer's rule for example; solve whichever you way you think is best. For brevity's sake I won't add more to it here, if you think you could benefit from an explanation on that, just contact me.

Anyhow, solving that we see that $a = \frac{1}{13}$, $b = - \frac{18}{143}$ and $d = \frac{7}{143}$

From there we can just say

$I = \frac{1}{13} \int \frac{\mathrm{dx}}{x - 2} - \frac{18}{143} \int \frac{\mathrm{dx}}{x + 4} + \frac{7}{143} \int \frac{\mathrm{dx}}{x - 7}$

Which are easy integrals

$I = \frac{1}{13} \ln | x - 2 | - \frac{18}{143} \ln | x + 4 | + \frac{7}{143} \ln | x - 7 | + c$