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

Lösung

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

- \frac{1}{6} ln ∣ x + 3 ∣ + \frac{1}{6} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x + 2 ∣ + C

Schritte zur Lösung

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

Wende U-Substitution an

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

Ermittle den Partialbruch von

\frac{1}{u ( u - 3 ) ( u - 2 ) ( u - 1 )} : - \frac{1}{6 u} + \frac{1}{6 ( u - 3 )} - \frac{1}{2 ( u - 2 )} + \frac{1}{2 ( u - 1 )}

= \int - \frac{1}{6 u} + \frac{1}{6 ( u - 3 )} - \frac{1}{2 ( u - 2 )} + \frac{1}{2 ( u - 1 )} 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}{6 u} d u + \int \frac{1}{6 ( u - 3 )} d u - \int \frac{1}{2 ( u - 2 )} d u + \int \frac{1}{2 ( u - 1 )} d u

\int \frac{1}{6 u} d u = \frac{1}{6} ln ∣ u ∣

\int \frac{1}{6 ( u - 3 )} d u = \frac{1}{6} ln ∣ u - 3 ∣

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

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

= - \frac{1}{6} ln ∣ u ∣ + \frac{1}{6} ln ∣ u - 3 ∣ - \frac{1}{2} ln ∣ u - 2 ∣ + \frac{1}{2} ln ∣ u - 1 ∣

Setze in

u = x + 3

ein

= - \frac{1}{6} ln ∣ x + 3 ∣ + \frac{1}{6} ln ∣ x + 3 - 3 ∣ - \frac{1}{2} ln ∣ x + 3 - 2 ∣ + \frac{1}{2} ln ∣ x + 3 - 1 ∣

Vereinfache

- \frac{1}{6} ln ∣ x + 3 ∣ + \frac{1}{6} ln ∣ x + 3 - 3 ∣ - \frac{1}{2} ln ∣ x + 3 - 2 ∣ + \frac{1}{2} ln ∣ x + 3 - 1 ∣ : - \frac{1}{6} ln ∣ x + 3 ∣ + \frac{1}{6} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x + 2 ∣

= - \frac{1}{6} ln ∣ x + 3 ∣ + \frac{1}{6} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x + 2 ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{6} ln ∣ x + 3 ∣ + \frac{1}{6} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x + 2 ∣ + C

\int \frac{e ^{3 x} + 3 e ^{x} + 3}{e ^{x}} d x

\int \frac{( y - 3 )}{4 y} d y

\int 5 x^{- 3} + 8 x^{- 2} + 2 x^{- 1} d x

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

\int e^{- x} sin (x) + e^{- x} cos (x) d x