# How do you integrate #int (x-5)/(x-2)^2# using partial fractions?

Partial fractions are not really necessary in this case because we can write: to divide the numerator.

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To integrate ( \frac{{x - 5}}{{(x - 2)^2}} ) using partial fractions, first express it as a sum of partial fractions:

[ \frac{{x - 5}}{{(x - 2)^2}} = \frac{A}{{x - 2}} + \frac{B}{{(x - 2)^2}} ]

Now, multiply both sides by ( (x - 2)^2 ) to clear the denominators:

[ x - 5 = A(x - 2) + B ]

Expand and equate coefficients:

[ x - 5 = Ax - 2A + B ]

From here, you can equate coefficients of like terms. Start by equating the constant terms:

[ -5 = -2A + B ]

Then equate the coefficients of ( x ):

[ 1 = A ]

Now that you have found the values of ( A ) and ( B ), substitute them back into the partial fractions:

[ \frac{{x - 5}}{{(x - 2)^2}} = \frac{1}{{x - 2}} + \frac{B}{{(x - 2)^2}} ]

[ \frac{{x - 5}}{{(x - 2)^2}} = \frac{1}{{x - 2}} + \frac{4}{{(x - 2)^2}} ]

Now, integrate each term separately:

[ \int \frac{{x - 5}}{{(x - 2)^2}} , dx = \int \frac{1}{{x - 2}} , dx + \int \frac{4}{{(x - 2)^2}} , dx ]

[ = \ln|x - 2| - \frac{4}{{x - 2}} + C ]

Where ( C ) is the constant of integration.

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