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.