How do you integrate $(3x) / (x-3)^2$ using partial fractions?

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Answer 1 · 2016-11-02T21:13:45

The answer is $=-9/(x-1)+3ln(x-3)+C$

Let's start the decomposition in partial fractions
$(3x)/(x-3)^2=A/(x-3)^2+B/(x-3)$

$=(A+B(x-3))/(x-3)^2$

$:.$ $3x=A+B(x-3)$
coefficients of x, $B=3$
$0=A-3B$ $=>$ $A=9$

$(3x)/(x-3)^2=9/(x-3)^2+3/(x-3)$

$int(3xdx)/(x-3)^2=int(9dx)/(x-3)^2+int(3dx)/(x-3)$

$=-9/(x-1)+3ln(x-3)+C$

How do you integrate (3x) / (x-3)^2(3x) / (x-3)^2 using partial fractions?