# How do you integrate int ((x-3)(x-1))/(-x(x-5)) dx using partial fractions?

May 18, 2018

$I = - x + \frac{3}{5} \ln | x | - \frac{8}{5} \ln | x - 5 | + C$

#### Explanation:

Here,

$I = \int \frac{\left(x - 3\right) \left(x - 1\right)}{- x \left(x - 5\right)} \mathrm{dx}$

=-int(x^2-4x+3)/(x^2-5x

$= - \int \frac{{x}^{2} - 5 x + x + 3}{{x}^{2} - 5 x}$

$= - \int \frac{{x}^{2} - 5 x}{{x}^{2} - 5 x} \mathrm{dx} - \int \frac{x + 3}{{x}^{2} - 5 x} \mathrm{dx}$

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

$I = - x - {I}_{1.} . . \to \left(D\right)$

We try to obtain Partial Fractions for ${I}_{1}$

$\frac{x + 3}{x \left(x - 5\right)} = \frac{A}{x} + \frac{B}{x - 5}$

$x + 3 = A \left(x - 5\right) + B x$

$x = 0 \implies 3 = A \left(- 5\right) \implies A = - \frac{3}{5}$

$x = 5 \implies 5 + 3 = 5 B \implies B = \frac{8}{5}$

So

${I}_{1} = \int \frac{- \frac{3}{5}}{x} \mathrm{dx} + \int \frac{\frac{8}{5}}{x - 5} \mathrm{dx}$

${I}_{1} = - \frac{3}{5} \int \frac{1}{x} \mathrm{dx} + \frac{8}{5} \int \frac{1}{x - 5} \mathrm{dx}$

${I}_{1} = - \frac{3}{5} \ln | x | + \frac{8}{5} \ln | x - 5 | + c$

Thus, from $\left(D\right)$,we get

$I = - x - \left\{- \frac{3}{5} \ln | x | + \frac{8}{5} \ln | x - 5 |\right\} + C$

$I = - x + \frac{3}{5} \ln | x | - \frac{8}{5} \ln | x - 5 | + C$