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

Lösung

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

\frac{3}{16} ln ∣ x ∣ + \frac{3}{16 x} - \frac{1}{16 x ^{2}} - \frac{3}{16} ln ∣ x + 2 ∣ + \frac{3}{16 ( x + 2 )} + \frac{1}{16 ( x + 2 ) ^{2}} + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{1}{x ^{3} ( x + 2 ) ^{3}} : \frac{3}{16 x} - \frac{3}{16 x ^{2}} + \frac{1}{8 x ^{3}} - \frac{3}{16 ( x + 2 )} - \frac{3}{16 ( x + 2 ) ^{2}} - \frac{1}{8 ( x + 2 ) ^{3}}

= \int \frac{3}{16 x} - \frac{3}{16 x ^{2}} + \frac{1}{8 x ^{3}} - \frac{3}{16 ( x + 2 )} - \frac{3}{16 ( x + 2 ) ^{2}} - \frac{1}{8 ( x + 2 ) ^{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{3}{16 x} d x - \int \frac{3}{16 x ^{2}} d x + \int \frac{1}{8 x ^{3}} d x - \int \frac{3}{16 ( x + 2 )} d x - \int \frac{3}{16 ( x + 2 ) ^{2}} d x - \int \frac{1}{8 ( x + 2 ) ^{3}} d x

\int \frac{3}{16 x} d x = \frac{3}{16} ln ∣ x ∣

\int \frac{3}{16 x ^{2}} d x = - \frac{3}{16 x}

\int \frac{1}{8 x ^{3}} d x = - \frac{1}{16 x ^{2}}

\int \frac{3}{16 ( x + 2 )} d x = \frac{3}{16} ln ∣ x + 2 ∣

\int \frac{3}{16 ( x + 2 ) ^{2}} d x = - \frac{3}{16 ( x + 2 )}

\int \frac{1}{8 ( x + 2 ) ^{3}} d x = - \frac{1}{16 ( x + 2 ) ^{2}}

= \frac{3}{16} ln ∣ x ∣ - (- \frac{3}{16 x}) - \frac{1}{16 x ^{2}} - \frac{3}{16} ln ∣ x + 2 ∣ - (- \frac{3}{16 ( x + 2 )}) - (- \frac{1}{16 ( x + 2 ) ^{2}})

Vereinfache

= \frac{3}{16} ln ∣ x ∣ + \frac{3}{16 x} - \frac{1}{16 x ^{2}} - \frac{3}{16} ln ∣ x + 2 ∣ + \frac{3}{16 ( x + 2 )} + \frac{1}{16 ( x + 2 ) ^{2}}

Füge eine Konstante zur Lösung hinzu

= \frac{3}{16} ln ∣ x ∣ + \frac{3}{16 x} - \frac{1}{16 x ^{2}} - \frac{3}{16} ln ∣ x + 2 ∣ + \frac{3}{16 ( x + 2 )} + \frac{1}{16 ( x + 2 ) ^{2}} + C

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