integral of (6x^4+x-3)/(2x^3+2x^2-12x)

Lösung

\int \frac{6 x ^{4} + x - 3}{2 x ^{3} + 2 x ^{2} - 12 x} d x

\frac{3 x ^{2}}{2} - 3 x + \frac{1}{4} ln ∣ x ∣ + \frac{19}{4} ln ∣ x - 2 ∣ + 16 ln ∣ x + 3 ∣ + C

Schritte zur Lösung

\int \frac{6 x ^{4} + x - 3}{2 x ^{3} + 2 x ^{2} - 12 x} d x

Ermittle den Partialbruch von

\frac{6 x ^{4} + x - 3}{2 x ^{3} + 2 x ^{2} - 12 x} : 3 x - 3 + \frac{1}{4 x} + \frac{19}{4 ( x - 2 )} + \frac{16}{x + 3}

= \int 3 x - 3 + \frac{1}{4 x} + \frac{19}{4 ( x - 2 )} + \frac{16}{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 3 x d x - \int 3 d x + \int \frac{1}{4 x} d x + \int \frac{19}{4 ( x - 2 )} d x + \int \frac{16}{x + 3} d x

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

\int 3 d x = 3 x

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

\int \frac{19}{4 ( x - 2 )} d x = \frac{19}{4} ln ∣ x - 2 ∣

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

= \frac{3 x ^{2}}{2} - 3 x + \frac{1}{4} ln ∣ x ∣ + \frac{19}{4} ln ∣ x - 2 ∣ + 16 ln ∣ x + 3 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{3 x ^{2}}{2} - 3 x + \frac{1}{4} ln ∣ x ∣ + \frac{19}{4} ln ∣ x - 2 ∣ + 16 ln ∣ x + 3 ∣ + C

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