How do you integrate #(2x-3)/(x^2-5x+6)# using partial fractions?

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Answer 1 · 2016-12-03T02:18:54

Please see the explanation.

Do partial fraction decomposition:

#(2x - 3)/((x - 2)(x - 3)) = A/(x - 2) + B/(x - 3)#

Multiply both sides by #((x - 2)(x - 3))#

#(2x - 3) = A(x - 3) + B(x - 2)#

Solve for A by letting #x = 2#:

#(2(2) - 3) = A(2 - 3)#

#-1 = A#

Solve for B by letting #x = 3#:

#(2(3) - 3) = B(3 - 2)#

3 = B

Check:

#-1/(x - 2) + 3/(x - 3) = #

#-1/(x - 2)(x - 3)/(x - 3) + 3/(x - 3)(x - 2)/(x - 2) =#

#(-x + 3)/((x - 2)(x - 3)) + (3x - 6)/((x - 3)(x - 2)) =#

#(2x - 3)/((x - 3)(x - 2))#

This checks.

#int (2x - 3)/(x^2 - 5x + 6)dx = 3int 1/(x - 3)dx - int1/(x - 2)dx#

#int (2x - 3)/(x^2 - 5x + 6)dx = 3ln|x - 3| - ln|x - 2| + C#