integral of ((x-4))/(x^2-5x+6)

Lösung

\int \frac{( x - 4 )}{x ^{2} - 5 x + 6} d x

- ln ∣ x - 3 ∣ + 2 ln ∣ x - 2 ∣ + C

Schritte zur Lösung

\int \frac{x - 4}{x ^{2} - 5 x + 6} d x

Wende U-Substitution an

= \int \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 - \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 \frac{1}{u + 1} d u + \int \frac{2}{u + 2} d u

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

\int \frac{2}{u + 2} d u = 2 ln ∣ u + 2 ∣

= - ln ∣ u + 1 ∣ + 2 ln ∣ u + 2 ∣

Setze in

u = x - 4

ein

= - ln ∣ x - 4 + 1 ∣ + 2 ln ∣ x - 4 + 2 ∣

Vereinfache

= - ln ∣ x - 3 ∣ + 2 ln ∣ x - 2 ∣

Füge eine Konstante zur Lösung hinzu

= - ln ∣ x - 3 ∣ + 2 ln ∣ x - 2 ∣ + C

\int x^{4} e^{1 - x^{5}} d x

t an g e n t 2 x^{3} + 3 x^{2} - 12 x + 1

\frac{d}{d y} (y e^{x y} cos (2 x) - 2 e^{x y} sin (2 x) + 2 x)

n = 1 \sum \infty \frac{( n ! ) ^{2}}{1}

x \to a lim (3 x^{2} - 6 x)