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 -3ln(2)+ln(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

- 3 ln (2) + ln (3)

Decimal

- 0.98082 \dots

Solution steps

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

Apply u-substitution

= \int_{- 4}^{- 3} \frac{u}{u ^{2} + 3 u + 2} d u

Take the partial fraction of

\frac{u}{u ^{2} + 3 u + 2} : - \frac{1}{u + 1} + \frac{2}{u + 2}

= \int_{- 4}^{- 3} - \frac{1}{u + 1} + \frac{2}{u + 2} d u

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}^{- 3} \frac{1}{u + 1} d u + \int_{- 4}^{- 3} \frac{2}{u + 2} d u

\int_{- 4}^{- 3} \frac{1}{u + 1} d u = ln (2) - ln (3)

\int_{- 4}^{- 3} \frac{2}{u + 2} d u = - 2 ln (2)

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

Simplify

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

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 -3ln(2)+ln(3)