Question

How do you integrate `int (6(5 - x))/( (x - 7)(4 - x))` using partial fractions?

Answer 1

`int(6(5-x))/((x-7)(4-x)) dx=4ln|x-7|+2ln|x-4|+C`

Explanation:

By multiplying the numerator and the denominator by `-1`,

`(6(x-5))/((x-7)(x-4))=A/(x-7)+B/(x-4)=(A(x-4)+B(x-7))/((x-7)(x-4))`

By matching the numerators,

`A(x-4)+B(x-7)=6(x-5)`

To find `A`, set `x=7 Rightarrow 3A=12 Rightarrow A=4`

To find `B`, set `x=4 Rightarrow -3B=-6 Rightarrow B=2`

So, we have

#int(6(5-x))/((x-7)(4-x))dx =int(4/(x-7)+2/(x-4) )dx#

By Log Rule,

`=4ln|x-7|+2ln|x-4|+C`

I hope that this was clear.