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 Notation

- 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

= \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)

\int 2 x z + y^{2} d x

\int_{2}^{4} (x - 4)^{2} + 1 d x

\frac{d}{d x} (2 cos^{2} (x) - 1)

\int + x^{9} x^{5} + 5 d x

(2 + x) (\frac{d y}{d x}) - x y = 0

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)