# How do you integrate int (1-2x^2)/((x+9)(x+7)(x+1))  using partial fractions?

Feb 5, 2018

The denominator factors are linear and distinct.

#### Explanation:

This fact makes the problem a little easier.

$\frac{- 2 {x}^{2} + 1}{\left(x + 9\right) \left(x + 7\right) \left(x + 1\right)} = \frac{A}{x + 9} + \frac{B}{x + 7} + \frac{C}{x + 1}$

Multiply both sides by the denominator on the left side. After cancellation we have:

$- 2 {x}^{2} + 1 = A \left(x + 7\right) \left(x + 1\right) + B \left(x + 9\right) \left(x + 1\right) + C \left(x + 9\right) \left(x + 7\right)$

Short Cut #1:

The above is true for all values of x. In particular it is true for x = -1, which makes one of the factors zero. Let x = -1.

$- 2 {\left(- 1\right)}^{2} + 1 = C \left(- 1 + 9\right) \left(- 1 + 7\right)$
$- 1 = C \left(8\right) \left(6\right)$
$- 1 = 48 C$

So $C = - \frac{1}{48}$.

Try the same short cut with x = -7 and then with x = -9, and you will have the values of A, B, and C.

Put those values into $\frac{A}{x + 9} + \frac{B}{x + 7} + \frac{C}{x + 1}$ and integrate term by term. You will get three natural logarithms if you do it right.