How do you integrate $\frac{2 x - 3}{x^{2} - 5 x + 6}$ $(2x-3)/(x^2-5x+6)$ using partial fractions?

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How do you integrate $\frac{2 x - 3}{x^{2} - 5 x + 6}$ $(2x-3)/(x^2-5x+6)$ using partial fractions?

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Answer 1 · 2016-12-03T02:18:54

Please see the explanation.

Explanation:

Do partial fraction decomposition:

$\frac{2 x - 3}{(x - 2) (x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}$ $(2x - 3)/((x - 2)(x - 3)) = A/(x - 2) + B/(x - 3)$

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

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

Solve for A by letting $x = 2$ $x = 2$ :

$(2 (2) - 3) = A (2 - 3)$ $(2(2) - 3) = A(2 - 3)$

$- 1 = A$ $-1 = A$

Solve for B by letting $x = 3$ $x = 3$ :

$(2 (3) - 3) = B (3 - 2)$ $(2(3) - 3) = B(3 - 2)$

3 = B

Check:

$- \frac{1}{x - 2} + \frac{3}{x - 3} =$ $-1/(x - 2) + 3/(x - 3) =$

$- \frac{1}{x - 2} \frac{x - 3}{x - 3} + \frac{3}{x - 3} \frac{x - 2}{x - 2} =$ $-1/(x - 2)(x - 3)/(x - 3) + 3/(x - 3)(x - 2)/(x - 2) =$

$\frac{- x + 3}{(x - 2) (x - 3)} + \frac{3 x - 6}{(x - 3) (x - 2)} =$ $(-x + 3)/((x - 2)(x - 3)) + (3x - 6)/((x - 3)(x - 2)) =$

$\frac{2 x - 3}{(x - 3) (x - 2)}$ $(2x - 3)/((x - 3)(x - 2))$

This checks.

$\int \frac{2 x - 3}{x^{2} - 5 x + 6} d x = 3 \int \frac{1}{x - 3} d x - \int \frac{1}{x - 2} d x$ $int (2x - 3)/(x^2 - 5x + 6)dx = 3int 1/(x - 3)dx - int1/(x - 2)dx$

$\int \frac{2 x - 3}{x^{2} - 5 x + 6} d x = 3 ln | x - 3 | - ln | x - 2 | + C$ $int (2x - 3)/(x^2 - 5x + 6)dx = 3ln|x - 3| - ln|x - 2| + C$

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How do you integrate $\frac{2 x - 3}{x^{2} - 5 x + 6}$ $(2x-3)/(x^2-5x+6)$ using partial fractions?

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Explanation:

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How do you integrate $\frac{2 x - 3}{x^{2} - 5 x + 6}$ $(2x-3)/(x^2-5x+6)$ using partial fractions?