# How do you integrate int (x-9)/((x+3)(x-7)(x-5))  using partial fractions?

Feb 8, 2016

$- \frac{3}{20} \ln \left\mid x + 3 \right\mid - \frac{1}{10} \ln \left\mid x - 7 \right\mid + \frac{1}{4} \ln \left\mid x - 5 \right\mid + C$

#### Explanation:

As the other answer was unfortunately deleted, I will answer this question.

To compute a partial fractions decomposition means to find $A$, $B$ and $C$ such as

$\frac{x - 9}{\left(x + 3\right) \left(x - 7\right) \left(x - 5\right)} = \frac{A}{x + 3} + \frac{B}{x - 7} + \frac{C}{x - 5}$

... multiply both sides with the denominator $\left(x + 3\right) \left(x - 7\right) \left(x - 5\right)$...

$\iff \text{ } x - 9 = A \left(x - 7\right) \left(x - 5\right) + B \left(x + 3\right) \left(x - 5\right) + C \left(x + 3\right) \left(x - 7\right)$

$\iff \text{ } x - 9 = A \left({x}^{2} - 12 x + 35\right) + B \left({x}^{2} - 2 x - 15\right) + C \left({x}^{2} - 4 x - 21\right)$

$\iff \text{ } x - 9 = A \cdot {x}^{2} - A \cdot 12 x + A \cdot 35 + B \cdot {x}^{2} - B \cdot 2 x - B \cdot 15 + C \cdot {x}^{2} - C \cdot 4 x - C \cdot 21$

... identify the $\textcolor{red}{{x}^{2}}$ terms, the $\textcolor{b l u e}{x}$ terms and the $\text{constant}$ terms:

$\iff \text{ } \textcolor{red}{0 \cdot {x}^{2}} + \textcolor{b l u e}{x} - 9 = \textcolor{red}{A \cdot {x}^{2}} \textcolor{b l u e}{- 12 A \cdot x} + 35 A + \textcolor{red}{B \cdot {x}^{2}} \textcolor{b l u e}{- 2 B \cdot x} - 15 B + \textcolor{red}{C \cdot {x}^{2}} \textcolor{b l u e}{- 4 C \cdot x} - 21 C$

This leads us to the following linear equation system:

{ (" " (I) " "0 = " "A + B " "+C " " color(red)(x^2) " terms"), (" " (II) " "1 = -12A - 2 B " "- 4C " " color(blue)(x) " terms"), (" " (III) -9 = " "35A - 15B - 21C " constant terms") :}

Now, let's solve this linear equation system.

To eliminate $B$, let's compute $\left(I V\right) = 2 \cdot \left(I\right) + \left(I I\right)$ and $\left(V\right) = 15 \cdot \left(I\right) + \left(I I I\right)$:

$\left(I V\right) \text{ } 1 = - 10 A - 2 C$
$\left(V\right) \text{ " -9 = " } 50 A - 6 C$

Now, we can compute $\left(V I\right) = 5 \cdot \left(I V\right) + \left(V\right)$ to eliminate $A$:

$\left(V I\right) \text{ } - 4 = - 16 C$

Thus, we can solve $\left(V I\right)$ for $C$ and find $C = \frac{1}{4}$.

Plugging $C$ into $\left(I V\right)$ gives us $A = - \frac{3}{20}$ and finally, plugging $A$ and $C$ into $\left(I\right)$ gives us $B = - \frac{1}{10}$.

Now, we can perform the partial fraction decomposition and solve the integral:

$\int \frac{x - 9}{\left(x + 3\right) \left(x - 7\right) \left(x - 5\right)} \text{d} x$

$= \int \left[- \frac{3}{20} \cdot \frac{1}{x + 3} - \frac{1}{10} \cdot \frac{1}{x - 7} + \frac{1}{4} \cdot \frac{1}{x - 5}\right] \text{d} x$

= -3/20 int [ 1 / (x+3)"d"x -1/10 int 1 / (x-7) "d"x + 1/4 int 1/ (x-5) "d"x

$= - \frac{3}{20} \ln \left\mid x + 3 \right\mid - \frac{1}{10} \ln \left\mid x - 7 \right\mid + \frac{1}{4} \ln \left\mid x - 5 \right\mid + C$