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

1 Answer
Feb 29, 2016

Answer:

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

Explanation:

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.