integral of 1/(x(x+1)(x+2))

Lösung

\int \frac{1}{x ( x + 1 ) ( x + 2 )} d x

\frac{1}{2} ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x + 2 ∣ + C

Schritte zur Lösung

\int \frac{1}{x ( x + 1 ) ( x + 2 )} d x

Ermittle den Partialbruch von

\frac{1}{x ( x + 1 ) ( x + 2 )} : \frac{1}{2 x} - \frac{1}{x + 1} + \frac{1}{2 ( x + 2 )}

= \int \frac{1}{2 x} - \frac{1}{x + 1} + \frac{1}{2 ( x + 2 )} d x

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}{2 x} d x - \int \frac{1}{x + 1} d x + \int \frac{1}{2 ( x + 2 )} d x

\int \frac{1}{2 x} d x = \frac{1}{2} ln ∣ x ∣

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

\int \frac{1}{2 ( x + 2 )} d x = \frac{1}{2} ln ∣ x + 2 ∣

= \frac{1}{2} ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x + 2 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x + 2 ∣ + C

\frac{d}{d x} (e^{\frac{- x}{4}})

\frac{\partial}{\partial y} (e^{- x} cos (yz) z)

\frac{\partial}{\partial y} (3 x - 4 x y + 9 y)

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

\int_{0}^{1} 5 x - 5 x^{2} d x