# How do you integrate int (x-5) / (x^2(x+1)) using partial fractions?

Dec 21, 2016

$6 \ln | x | + \frac{5}{x} - 6 \ln | x + 1 | + C$

#### Explanation:

Decompose into partial fractions:
$\frac{x - 5}{{x}^{2} \left(x + 1\right)} \equiv \frac{A x + B}{x} ^ 2 - \frac{6}{x + 1}$
giving
$x - 5 \equiv \left(A x + B\right) \left(x + 1\right) - 6 {x}^{2}$
Regrouping:
$0 {x}^{2} + {x}^{1} - 5 {x}^{0} \equiv \left(A - 6\right) {x}^{2} + \left(A + B\right) {x}^{1} + B {x}^{0}$
Equating powers of $x$:
$A = 6$, $B = - 5$
giving the integrand as
$\frac{6 x - 5}{x} ^ 2 - \frac{6}{x + 1}$
$= \frac{6}{x} - \frac{5}{x} ^ 2 - \frac{6}{x + 1}$
The $6$ in the partial fractions comes from the cover-up rule, which avoids getting three simultaneous equations (one for each power of $x$ in the identity) to solve.