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)

Decimal

0.40546 \dots

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)

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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)