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.

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

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

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