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

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Answer 1 · 2017-04-13T18:41:53

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

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.

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