# How do you integrate int (1-2x^2)/((x-2)(x+6)(x+4))  using partial fractions?

$- \frac{7}{48} \ln \left(x - 2\right) - \frac{71}{16} \ln \left(x + 6\right) + \frac{31}{12} \ln \left(x + 4\right) + C$

#### Explanation:

from the given
$\int \frac{1 - 2 {x}^{2}}{\left(x - 2\right) \left(x + 6\right) \left(x + 4\right)}$ $\mathrm{dx}$

$\frac{A}{x - 2} + \frac{B}{x + 6} + \frac{C}{x + 4}$ = $\frac{1 - 2 {x}^{2}}{\left(x - 2\right) \left(x + 6\right) \left(x + 4\right)}$

then

$A \left(x + 6\right) \left(x + 4\right) + B \left(x - 2\right) \left(x + 4\right) + C \left(x - 2\right) \left(x + 6\right)$

= $1 - 2 {x}^{2}$

then

$A \left({x}^{2} + 10 x + 24\right) + B \left({x}^{2} + 2 x - 8\right) + C \left({x}^{2} + 4 x - 12\right)$

=$1 - 2 {x}^{2}$

The 3 equations are now defined

$A + B + C = - 2$

$10 A + 2 B + 4 C = 0$

$24 A - 8 B - 12 C = 1$

Use Algebra to find A, B, C

$A = - \frac{7}{48}$

$B = - \frac{71}{16}$

$C = \frac{31}{12}$

The integral becomes:

$\int \frac{- \frac{7}{48}}{x - 2}$$\mathrm{dx}$+$\int \frac{- \frac{71}{16}}{x + 6}$$\mathrm{dx}$+$\int \frac{\frac{31}{12}}{x + 4}$$\mathrm{dx}$

$- \frac{7}{48} \ln \left(x - 2\right)$ -$\frac{71}{16} \ln \left(x + 6\right)$+$\frac{31}{12} \ln \left(x + 4\right)$