What is the integral from 0 to 1 of (x-4)/(x^2+5x+6) ?

The integral from 0 to 1 of (x-4)/(x^2+5x+6) is 20ln(2)-13ln(3)

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integral from 0 to 1 of (x-4)/(x^2+5x+6)

\int_{0}^{1} \frac{x - 4}{x ^{2} + 5 x + 6} d x

20 ln (2) - 13 ln (3)

Decimal Notation

- 0.41901 \dots

Solution steps

\int_{0}^{1} \frac{x - 4}{x ^{2} + 5 x + 6} d x

Take the partial fraction of

\frac{x - 4}{x ^{2} + 5 x + 6} : - \frac{6}{x + 2} + \frac{7}{x + 3}

= \int_{0}^{1} - \frac{6}{x + 2} + \frac{7}{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_{0}^{1} \frac{6}{x + 2} d x + \int_{0}^{1} \frac{7}{x + 3} d x

\int_{0}^{1} \frac{6}{x + 2} d x = 6 (ln (3) - ln (2))

\int_{0}^{1} \frac{7}{x + 3} d x = 7 (2 ln (2) - ln (3))

= - 6 (ln (3) - ln (2)) + 7 (2 ln (2) - ln (3))

Simplify

- 6 (ln (3) - ln (2)) + 7 (2 ln (2) - ln (3)) : 20 ln (2) - 13 ln (3)

= 20 ln (2) - 13 ln (3)

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What is the integral from 0 to 1 of (x-4)/(x^2+5x+6) ?
The integral from 0 to 1 of (x-4)/(x^2+5x+6) is 20ln(2)-13ln(3)