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

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Answer 1 · 2016-02-29T21:27:14

$int x/(x-3 )^2dx = ln(|x-3|) - 3/(x-3) + C$ , where $C$ is the constant of integration.

Using the partial fractions decomposition, we write $x/(x-3 )^2$ in the form of $A/(x-3) + B/(x-3)^2$ , which is easier to integrate.

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

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

$iff A = 1 and -3A + B = 0 iff A = 1 and B = 3.$

Therefore, $x/(x-3 )^2 = 1/(x-3) + 3/(x-3)^2.$

And $int x/(x-3 )^2dx = int 1/(x-3) dx + 3* int 1/(x-3)^2 dx$

$int x/(x-3 )^2dx = ln(|x-3|) - 3/(x-3) + C$ , where $C$ is the constant of integration.

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