partial fractions

[missyc8]

integral (3x^2 - 10/ x^2 -4x+ 4) dx

bottom = (x-2) (x-2)

A/(x-2) + B/ (x-2)^2

where do i go from here? i do not understand...thanks!

 

[Deleted member 4993]

integral (3x^2 - 10/ x^2 -4x+ 4) dx

bottom = (x-2) (x-2)

A/(x-2) + B/ (x-2)^2

where do i go from here? i do not understand...thanks!

First you need to write it in a way that numerator is lower degree polynomial than denominator (right now those are both 2nd degree polynomial)

\(\displaystyle \frac{3x^2 - 10}{x^2 - 4x + 4} \, = \, 3 \, + \, 2\cdot \frac{6x - 11}{x^2 - 4x + 4}\)

Then

\(\displaystyle \frac{6x - 11}{x^2 - 4x + 4} \, = \, \frac{A}{x-2} \, + \, \frac{B}{(x-2)^2}\)

\(\displaystyle \frac{6x - 11}{x^2 - 4x + 4} \, = \, \frac{A(x-2) + B}{(x-2)^2}\)

Now solve for A & B by equating the numerators