What is the integral from 0 to 1 of (8x+1)/(x^2-x-6) ?

The integral from 0 to 1 of (8x+1)/(x^2-x-6) is 2ln(2)-2ln(3)

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integral from 0 to 1 of (8x+1)/(x^2-x-6)

\int_{0}^{1} \frac{8 x + 1}{x ^{2} - x - 6} d x

2 ln (2) - 2 ln (3)

Decimal Notation

- 0.81093 \dots

Solution steps

\int_{0}^{1} \frac{8 x + 1}{x ^{2} - x - 6} d x

Take the partial fraction of

\frac{8 x + 1}{x ^{2} - x - 6} : \frac{3}{x + 2} + \frac{5}{x - 3}

= \int_{0}^{1} \frac{3}{x + 2} + \frac{5}{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{3}{x + 2} d x + \int_{0}^{1} \frac{5}{x - 3} d x

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

\int_{0}^{1} \frac{5}{x - 3} d x = 5 (ln (2) - ln (3))

= 3 (ln (3) - ln (2)) + 5 (ln (2) - ln (3))

Simplify

3 (ln (3) - ln (2)) + 5 (ln (2) - ln (3)) : 2 ln (2) - 2 ln (3)

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

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What is the integral from 0 to 1 of (8x+1)/(x^2-x-6) ?
The integral from 0 to 1 of (8x+1)/(x^2-x-6) is 2ln(2)-2ln(3)