How do you integrate $int (2x-2)/((x-4)(x-1)(x-6))$ using partial fractions?

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Answer 1 · 2016-11-15T08:19:36

The answer is $=-ln(x-4)+ln(x-6) +C$

Let's start the decomposition into partial fractions

$(2x-2)/((x-4)(x-1)(x-6))=(2cancel(x-1))/((x-4)cancel(x-1)(x-6))$

$=2/((x-4)(x-6))=A/(x-4)+B/(x-6)$

$=(A(x-6)+B(x-4))/((x-4)(x-6))$

So, $2=(A(x-6)+B(x-4))$

Let $x=4$ $=>$ $2=-2A$ $=>$ $A=-1$

Let $x=6$ $=>$ $2=2B$ $=>$ $B=1$

$2/((x-4)(x-6))=-1/(x-4)+1/(x-6)$

$int(2dx)/((x-4)(x-6))=int(-1dx)/(x-4)+int(1dx)/(x-6)$

$=-ln(x-4)+ln(x-6) +C$

How do you integrate int (2x-2)/((x-4)(x-1)(x-6)) int (2x-2)/((x-4)(x-1)(x-6)) using partial fractions?