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

Lösung

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

- 3 ln (2) + ln (3)

Dezimale

- 0.98082 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Ermittle den Partialbruch von

\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

Wende die Summenregel an:

\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)

Vereinfache

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

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

(1 - x^{2}) y^{^{'}} + x y = x

\int \frac{2 x}{k ^{2}} d x

\frac{d}{d x} (x + \frac{25}{x})

\int x^{3} ln (8 x) d x