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

Dec 27, 2016

The answer is =-4/7ln(∣x-5∣)-3/7ln(∣x+2∣)+ C

#### Explanation:

$1 - {x}^{2} = \left(1 + x\right) \left(1 - x\right)$

Let's rewrite the expression

$\frac{1 - {x}^{2}}{\left(x + 1\right) \left(x - 5\right) \left(x + 2\right)} = \frac{\cancel{1 + x} \left(1 - x\right)}{\cancel{x + 1} \left(x - 5\right) \left(x + 2\right)}$

Now we can do the decomposition into partial fractions

$\frac{1 - x}{\left(x - 5\right) \left(x + 2\right)} = \frac{A}{x - 5} + \frac{B}{x + 2}$

$= \frac{A \left(x + 2\right) + B \left(x - 5\right)}{\left(x - 5\right) \left(x + 2\right)}$

Therefore,

$1 - x = A \left(x + 2\right) + B \left(x - 5\right)$

Let $x = - 2$, $\implies$, $3 = - 7 B$, $\implies$, $B = - \frac{3}{7}$

Let $x = 5$, $\implies$, $- 4 = 7 A$, $\implies$, $A = - \frac{4}{7}$

So,

$\frac{1 - x}{\left(x - 5\right) \left(x + 2\right)} = \frac{- \frac{4}{7}}{x - 5} + \frac{- \frac{3}{7}}{x + 2}$

Therefore,

int((1-x)dx)/((x-5)(x+2))=int((-4/7)dx)/(x-5)+int((-3/7)dx)/(x+2

=-4/7ln(∣x-5∣)-3/7ln(∣x+2∣)+ C