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

Jan 13, 2018

THe answer is $= - \frac{5}{8} \ln \left(| x - 6 |\right) - \frac{3}{8} \ln \left(| x + 2 |\right) + C$

#### Explanation:

Perform the decomposition into partial fractions

$\frac{1 - {x}^{2}}{\left(x + 1\right) \left(x - 6\right) \left(x + 2\right)} = \frac{A}{x + 1} + \frac{B}{x - 6} + \frac{C}{x + 2}$

$= \frac{A \left(x - 6\right) \left(x + 2\right) + B \left(x + 1\right) \left(x + 2\right) + C \left(x + 1\right) \left(x - 6\right)}{\left(x + 1\right) \left(x - 6\right) \left(x + 2\right)}$

The denominators are the same, compare the numerators

$1 - {x}^{2} = A \left(x - 6\right) \left(x + 2\right) + B \left(x + 1\right) \left(x + 2\right) + C \left(x + 1\right) \left(x - 6\right)$

Let $x = - 1$, $\implies$, $0 = A$

Let $x = 6$, $\implies$, $- 35 = B \cdot 7 \cdot 8$, $\implies$, $B = - \frac{5}{8}$

Let $x = - 2$, $\implies$, $- 3 = C \cdot - 1 \cdot - 8$, $C = - \frac{3}{8}$

Therefore,

$\frac{1 - {x}^{2}}{\left(x + 1\right) \left(x - 6\right) \left(x + 2\right)} = \frac{0}{x + 1} + \frac{- \frac{5}{8}}{x - 6} + \frac{- \frac{3}{8}}{x + 2}$

So,

$\int \frac{\left(1 - {x}^{2}\right) \mathrm{dx}}{\left(x + 1\right) \left(x - 6\right) \left(x + 2\right)} = \int \frac{- \frac{5}{8} \mathrm{dx}}{x - 6} + \int \frac{- \frac{3}{8} \mathrm{dx}}{x + 2}$

$= - \frac{5}{8} \ln \left(| x - 6 |\right) - \frac{3}{8} \ln \left(| x + 2 |\right) + C$

Jan 13, 2018

$\int \frac{1 - {x}^{2}}{\left(x + 1\right) \left(x - 6\right) \left(x + 2\right)} \cdot \mathrm{dx}$

=$- \frac{3}{8} L n \left(x + 2\right) - \frac{5}{8} L n \left(x - 6\right) + C$

#### Explanation:

$\int \frac{1 - {x}^{2}}{\left(x + 1\right) \left(x - 6\right) \left(x + 2\right)} \cdot \mathrm{dx}$

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

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

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

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

Now I decomposed integrand,

$\frac{1 - x}{\left(x - 6\right) \cdot \left(x + 2\right)} = \frac{A}{x + 2} + \frac{B}{x - 6}$

After expanding denominator,

$A \cdot \left(x - 6\right) + B \cdot \left(x + 2\right) = 1 - x$

Set $x = - 2$, $- 8 A = 3$. So $A = - \frac{3}{8}$

Set $x = 6$, $8 B = - 5$. So $B = - \frac{5}{8}$

Hence,

$\int \frac{1 - x}{\left(x - 6\right) \cdot \left(x + 2\right)} \cdot \mathrm{dx}$

=$- \frac{3}{8} \int \frac{\mathrm{dx}}{x + 2} - \frac{5}{8} \int \frac{\mathrm{dx}}{x - 6}$

=$- \frac{3}{8} L n \left(x + 2\right) - \frac{5}{8} L n \left(x - 6\right) + C$