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,

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

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

$=-1/2*ln(∣3x-5∣)/3+1/2*ln(∣x-3∣)+C$

$=-1/6*ln(∣3x-5∣)+1/2*ln(∣x-3∣)+C$

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