What is the integral from 4 to 6 of (3x-4)/(x^2-x-6) ?

The integral from 4 to 6 of (3x-4)/(x^2-x-6) is 2ln(4/3)+ln(3)

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integral from 4 to 6 of (3x-4)/(x^2-x-6)

\int_{4}^{6} \frac{3 x - 4}{x ^{2} - x - 6} d x

2 ln (\frac{4}{3}) + ln (3)

Decimal

1.67397 \dots

Solution steps

\int_{4}^{6} \frac{3 x - 4}{x ^{2} - x - 6} d x

Take the partial fraction of

\frac{3 x - 4}{x ^{2} - x - 6} : \frac{2}{x + 2} + \frac{1}{x - 3}

= \int_{4}^{6} \frac{2}{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_{4}^{6} \frac{2}{x + 2} d x + \int_{4}^{6} \frac{1}{x - 3} d x

\int_{4}^{6} \frac{2}{x + 2} d x = 2 ln (\frac{4}{3})

\int_{4}^{6} \frac{1}{x - 3} d x = ln (3)

= 2 ln (\frac{4}{3}) + ln (3)

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