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

Answer 1

#ln|x-2|+3/(x-2)+C#

#int(x-5)/(x-2)^2dx# #=int(x-2-3)/(x-2)^2dx# #=int(cancel(x-2)/(x-2)^cancel2-3/(x-2)^2)dx# #=int(1/(x-2)-3/(x-2)^2)dx# #=ln|x-2|+3/(x-2)+C#.

If the last step is not obvious, substitute #u=x-2#. Alternatively, substitute #u=x-2# right at the start to get #int(1/u-3/u^2)du#, but this isn't then really partial fractions. However, it isn't really a partial fractions question because you have only one unique term in the denominator.

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Answer 2

Christopher Agresta

To integrate (x-5)/(x-2)^2 using partial fractions, follow these steps:

Perform polynomial long division to express the fraction as a sum of a polynomial and a proper rational function.
Write the fraction in the form A/(x-2) + B/(x-2)^2.
Multiply both sides of the equation by (x-2)^2 to clear the denominators.
Simplify the resulting equation and solve for the constants A and B by comparing coefficients.
Integrate each term separately.
Combine the integrals to get the final result.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.