Question

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

Answer 1

The answer is `=-2/3ln(|x+1|)+2/3ln(|x-5|)+C`

Explanation:

Let's perform the decomposition into partial fractions

`4/((x+1)(x-5))=A/(x+1)+B/(x-5)`

`=(A(x-5)+B(x+1))/((x+1)(x-5))`

As the denominators are the same, we compare the numerators

`4=A(x-5)+B(x+1)`

Let `x=-1`, `=>`, `4=-6A`, `=>`, `A=-2/3`

Let `x=5`, `=>`, `4=6B`, `=>`, `B=2/3`

Therefore,

`4/((x+1)(x-5))=(-2/3)/(x+1)+(2/3)/(x-5)`

So,

`int(4dx)/((x+1)(x-5))=-2/3intdx/(x+1)+2/3intdx/(x-5)`

`=-2/3ln(|x+1|)+2/3ln(|x-5|)+C`

`=2/3ln((|x-5|)/(|x+1|))+C`

Answer 2

`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 `4/((x+1)(x-5))=A/(x+1)+B/(x-5)`
then
`color(white)("XXX")A(x-5)+B(x+1)=4`

`color(white)("XXX")rarr Ax+Bx = 0x color(white)("XX")rarr A+B=0color(white)("XX")rarr A=-B`
and
`color(white)("XXX")rarr -5A+B=4`
Combining:
`color(white)("XXX")-5(-B)+B=4color(white)("XX")B=2/3color(white)("XX")rarr A=-2/3`

`4/((x+1)(x-5))=-2/3(1/(x+1))+2/3(1/(x-5))`

Part 2: Integration Using the Fractional Decompostion
`int 4/((x+1)(x-5)) dx`

`color(white)("XXX")=intcolor(white)("x")[-2/3(1/(x+1))+2/3(1/(x-5))]dx`

`color(white)("XXX")=-2/3int1/(x+1)dx+2/3int1/(x-5) dx`

`color(white)("XXXXXXXX")`Somewhere around here, it would be convenient to remember:
`color(white)("XXXXXXXXXXX")int 1/(x+a) dx = lnabs(x+a) +C`
`color(white)("XXXXXXXXXXX")`so we can continue...

`color(white)("XXX")=-2/3lnabs(x+1)+2/3lnabs(x-5)+C`

`color(white)("XXX")=2/3[lnabs(x-5)-lnabs(x+1)]+C`

`color(white)("XXX")=2/3ln((abs(x-5))/(abs(x+1))) +C`