What is the integral from-1 to 1 of 1/(x^2-5x+6) ?

The integral from-1 to 1 of 1/(x^2-5x+6) is ln(3)-ln(2)

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integral from-1 to 1 of 1/(x^2-5x+6)

\int_{- 1}^{1} \frac{1}{x ^{2} - 5 x + 6} d x

ln (3) - ln (2)

Solution steps

\int_{- 1}^{1} \frac{1}{x ^{2} - 5 x + 6} d x

Take the partial fraction of

\frac{1}{x ^{2} - 5 x + 6} : - \frac{1}{x - 2} + \frac{1}{x - 3}

= \int_{- 1}^{1} - \frac{1}{x - 2} + \frac{1}{x - 3} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{- 1}^{1} \frac{1}{x - 2} d x + \int_{- 1}^{1} \frac{1}{x - 3} d x

\int_{- 1}^{1} \frac{1}{x - 2} d x = - ln (3)

\int_{- 1}^{1} \frac{1}{x - 3} d x = - ln (2)

= - (- ln (3)) - ln (2)

Simplify

= ln (3) - ln (2)

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What is the integral from-1 to 1 of 1/(x^2-5x+6) ?
The integral from-1 to 1 of 1/(x^2-5x+6) is ln(3)-ln(2)