integral of 1/(x^2(4-x))

Lösung

\int \frac{1}{x ^{2} ( 4 - x )} d x

\frac{1}{16} ln ∣ x ∣ - \frac{1}{4 x} - \frac{1}{16} ln ∣ x - 4 ∣ + C

Schritte zur Lösung

\int \frac{1}{x ^{2} ( 4 - x )} d x

Ermittle den Partialbruch von

\frac{1}{x ^{2} ( 4 - x )} : \frac{1}{16 x} + \frac{1}{4 x ^{2}} - \frac{1}{16 ( x - 4 )}

= \int \frac{1}{16 x} + \frac{1}{4 x ^{2}} - \frac{1}{16 ( x - 4 )} 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}{16 x} d x + \int \frac{1}{4 x ^{2}} d x - \int \frac{1}{16 ( x - 4 )} d x

\int \frac{1}{16 x} d x = \frac{1}{16} ln ∣ x ∣

\int \frac{1}{4 x ^{2}} d x = - \frac{1}{4 x}

\int \frac{1}{16 ( x - 4 )} d x = \frac{1}{16} ln ∣ x - 4 ∣

= \frac{1}{16} ln ∣ x ∣ - \frac{1}{4 x} - \frac{1}{16} ln ∣ x - 4 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{16} ln ∣ x ∣ - \frac{1}{4 x} - \frac{1}{16} ln ∣ x - 4 ∣ + C