How do you integrate #int (x-1)/(3x^2-14x+15)# using partial fractions?

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Answer 1 · 2016-12-04T11:06:30

The answer is #=-1/6ln(∣3x-5∣)+1/2ln(∣x-3∣)+C#

Let's factorise the denominator

#3x^2-14x+15=(3x-5)(x-3)#

So, the decomposition into partial fractions are

#(x-1)/(3x^2-14x+15)=(x-1)/((3x-5)(x-3))=A/(3x-5)+B/(x-3)#

#=(A(x-3)+B(3x-5))/((3x-5)(x-3))#

So, #(x-1)=A(x-3)+B(3x-5)#

Let #x=3#, #=>#, #2=4B#, #=>#, #B=1/2#

Let #x=5/3#, #=>#, #2/3=-4A/3#, #=>#,#A=-1/2#

so,

#(x-1)/(3x^2-14x+15)=(-1/2)/(3x-5)+(1/2)/(x-3)#

#int((x-1)dx)/(3x^2-14x+15)=int((-1/2)dx)/(3x-5)+int((1/2)dx)/(x-3)#

#=-1/2*ln(∣3x-5∣)/3+1/2*ln(∣x-3∣)+C#

#=-1/6*ln(∣3x-5∣)+1/2*ln(∣x-3∣)+C#