integral of 1/((x-1)(x-2)(x-3))

Lösung

\int \frac{1}{( x - 1 ) ( x - 2 ) ( x - 3 )} d x

\frac{1}{2} ln ∣ x - 1 ∣ - ln ∣ x - 2 ∣ + \frac{1}{2} ln ∣ x - 3 ∣ + C

Schritte zur Lösung

\int \frac{1}{( x - 1 ) ( x - 2 ) ( x - 3 )} d x

Ermittle den Partialbruch von

\frac{1}{( x - 1 ) ( x - 2 ) ( x - 3 )} : \frac{1}{2 ( x - 1 )} - \frac{1}{x - 2} + \frac{1}{2 ( x - 3 )}

= \int \frac{1}{2 ( x - 1 )} - \frac{1}{x - 2} + \frac{1}{2 ( x - 3 )} 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 - 1 )} d x - \int \frac{1}{x - 2} d x + \int \frac{1}{2 ( x - 3 )} d x

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

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

\int \frac{1}{2 ( x - 3 )} d x = \frac{1}{2} ln ∣ x - 3 ∣

= \frac{1}{2} ln ∣ x - 1 ∣ - ln ∣ x - 2 ∣ + \frac{1}{2} ln ∣ x - 3 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln ∣ x - 1 ∣ - ln ∣ x - 2 ∣ + \frac{1}{2} ln ∣ x - 3 ∣ + C

\frac{d}{d x} (- \frac{30}{x ^{4}})

\int (1 - 2 x^{2} + 3 x^{3}) d x

l a pl a ce t r an s f or m e^{- t} \cdot sin (2 t)

f (x) = - sin (2 x) \cdot 2

\int \frac{x}{e ^{- x^{2}}} d x