# How do you integrate int (1-x^2)/((x+1)(x-6)(x-3))  using partial fractions?

Jul 17, 2018

The answer is $= - \frac{5}{3} \ln \left(| x - 6 |\right) + \frac{2}{3} \ln \left(| x - 3 |\right) + C$

#### Explanation:

First, simplify

$\frac{1 - {x}^{2}}{\left(x + 1\right) \left(x - 6\right) \left(x - 3\right)} = \frac{\left(1 - x\right) \cancel{1 + x}}{\cancel{x + 1} \left(x - 6\right) \left(x - 3\right)}$

Therefore,

$\frac{1 - x}{\left(x - 6\right) \left(x - 3\right)} = \frac{A}{x - 6} + \frac{B}{x - 3}$

$= \frac{A \left(x - 3\right) + B \left(x - 6\right)}{\left(x - 6\right) \left(x - 3\right)}$

The denominators are the same, compare the numerators

$1 - x = A \left(x - 3\right) + B \left(x - 6\right)$

Let $x = 6$, $- 5 = 3 A$, $\implies$, $A = - \frac{5}{3}$

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

Therefore,

$\frac{1 - x}{\left(x - 6\right) \left(x - 3\right)} = \frac{- \frac{5}{3}}{x - 6} + \frac{\frac{2}{3}}{x - 3}$

Therefore, the integral is

$I = \int \frac{\left(1 - x\right) \mathrm{dx}}{\left(x - 6\right) \left(x - 3\right)} = \int \frac{- \frac{5}{3} \mathrm{dx}}{x - 6} + \int \frac{\frac{2}{3} \mathrm{dx}}{x - 3}$

$= - \frac{5}{3} \ln \left(| x - 6 |\right) + \frac{2}{3} \ln \left(| x - 3 |\right) + C$

Jul 17, 2018

$\frac{2}{3} \ln | \left(x - 3\right) | - \frac{5}{3} \ln | \left(x - 6\right) | + C ,$

$\mathmr{and} , \ln | {\left(x - 3\right)}^{\frac{2}{3}} / {\left(x - 6\right)}^{\frac{5}{3}} | + C .$

#### Explanation:

We use the method of partial fraction in disguise.

Let, $I = \int \frac{1 - {x}^{2}}{\left(x + 1\right) \left(x - 6\right) \left(x - 3\right)} \mathrm{dx}$,

$= \int \frac{\left(1 + x\right) \left(1 - x\right)}{\left(x + 1\right) \left(x - 6\right) \left(x - 3\right)} \mathrm{dx}$.

$\therefore I = \int \frac{1 - x}{\left(x - 6\right) \left(x - 3\right)} \mathrm{dx}$,

Note that,

$\left(x - 3\right) - \left(x - 6\right) = 3 , \mathmr{and} , 2 \left(x - 3\right) - \left(x - 6\right) = x$.

$\therefore 1 - x = \frac{1}{3} \left(3\right) - x$,

$= \frac{1}{3} \left\{\left(x - 3\right) - \left(x - 6\right)\right\} - \left\{2 \left(x - 3\right) - \left(x - 6\right)\right\}$,

$= \left(\frac{1}{3} - 2\right) \left(x - 3\right) + \left(1 - \frac{1}{3}\right) \left(x - 6\right)$,

$= \frac{2}{3} \left(x - 6\right) - \frac{5}{3} \left(x - 3\right)$.

$\therefore I = \int \frac{\frac{2}{3} \left(x - 6\right) - \frac{5}{3} \left(x - 3\right)}{\left(x - 6\right) \left(x - 3\right)} \mathrm{dx}$,

$= \frac{2}{3} \int \frac{x - 6}{\left(x - 6\right) \left(x - 3\right)} \mathrm{dx} - \frac{5}{3} \int \frac{x - 3}{\left(x - 6\right) \left(x - 3\right)} \mathrm{dx}$,

$= \frac{2}{3} \int \frac{1}{x - 3} \mathrm{dx} - \frac{5}{3} \int \frac{1}{x - 6} \mathrm{dx}$,

$\Rightarrow I = \frac{2}{3} \ln | \left(x - 3\right) | - \frac{5}{3} \ln | \left(x - 6\right) | + C ,$

$\mathmr{and} , I = \ln | {\left(x - 3\right)}^{\frac{2}{3}} / {\left(x - 6\right)}^{\frac{5}{3}} | + C .$