# How do you integrate 4/((x+1)(x-5)) using partial fractions?

Feb 18, 2017

The answer is $= - \frac{2}{3} \ln \left(| x + 1 |\right) + \frac{2}{3} \ln \left(| x - 5 |\right) + C$

#### Explanation:

Let's perform the decomposition into partial fractions

$\frac{4}{\left(x + 1\right) \left(x - 5\right)} = \frac{A}{x + 1} + \frac{B}{x - 5}$

$= \frac{A \left(x - 5\right) + B \left(x + 1\right)}{\left(x + 1\right) \left(x - 5\right)}$

As the denominators are the same, we compare the numerators

$4 = A \left(x - 5\right) + B \left(x + 1\right)$

Let $x = - 1$, $\implies$, $4 = - 6 A$, $\implies$, $A = - \frac{2}{3}$

Let $x = 5$, $\implies$, $4 = 6 B$, $\implies$, $B = \frac{2}{3}$

Therefore,

$\frac{4}{\left(x + 1\right) \left(x - 5\right)} = \frac{- \frac{2}{3}}{x + 1} + \frac{\frac{2}{3}}{x - 5}$

So,

$\int \frac{4 \mathrm{dx}}{\left(x + 1\right) \left(x - 5\right)} = - \frac{2}{3} \int \frac{\mathrm{dx}}{x + 1} + \frac{2}{3} \int \frac{\mathrm{dx}}{x - 5}$

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

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

Feb 18, 2017

int 4/((x+1)(x-5)) dx=color(green)(2/3(ln((abs(x-5))/(abs(x+1))) +C)

#### Explanation:

Step 1: Partial Fraction Decomposition
If $\frac{4}{\left(x + 1\right) \left(x - 5\right)} = \frac{A}{x + 1} + \frac{B}{x - 5}$
then
$\textcolor{w h i t e}{\text{XXX}} A \left(x - 5\right) + B \left(x + 1\right) = 4$

$\textcolor{w h i t e}{\text{XXX")rarr Ax+Bx = 0x color(white)("XX")rarr A+B=0color(white)("XX}} \rightarrow A = - B$
and
$\textcolor{w h i t e}{\text{XXX}} \rightarrow - 5 A + B = 4$
Combining:
$\textcolor{w h i t e}{\text{XXX")-5(-B)+B=4color(white)("XX")B=2/3color(white)("XX}} \rightarrow A = - \frac{2}{3}$

$\frac{4}{\left(x + 1\right) \left(x - 5\right)} = - \frac{2}{3} \left(\frac{1}{x + 1}\right) + \frac{2}{3} \left(\frac{1}{x - 5}\right)$

Part 2: Integration Using the Fractional Decompostion
$\int \frac{4}{\left(x + 1\right) \left(x - 5\right)} \mathrm{dx}$

$\textcolor{w h i t e}{\text{XXX")=intcolor(white)("x}} \left[- \frac{2}{3} \left(\frac{1}{x + 1}\right) + \frac{2}{3} \left(\frac{1}{x - 5}\right)\right] \mathrm{dx}$

$\textcolor{w h i t e}{\text{XXX}} = - \frac{2}{3} \int \frac{1}{x + 1} \mathrm{dx} + \frac{2}{3} \int \frac{1}{x - 5} \mathrm{dx}$

$\textcolor{w h i t e}{\text{XXXXXXXX}}$Somewhere around here, it would be convenient to remember:
$\textcolor{w h i t e}{\text{XXXXXXXXXXX}} \int \frac{1}{x + a} \mathrm{dx} = \ln \left\mid x + a \right\mid + C$
$\textcolor{w h i t e}{\text{XXXXXXXXXXX}}$so we can continue...

$\textcolor{w h i t e}{\text{XXX}} = - \frac{2}{3} \ln \left\mid x + 1 \right\mid + \frac{2}{3} \ln \left\mid x - 5 \right\mid + C$

$\textcolor{w h i t e}{\text{XXX}} = \frac{2}{3} \left[\ln \left\mid x - 5 \right\mid - \ln \left\mid x + 1 \right\mid\right] + C$

$\textcolor{w h i t e}{\text{XXX}} = \frac{2}{3} \ln \left(\frac{\left\mid x - 5 \right\mid}{\left\mid x + 1 \right\mid}\right) + C$