integral of (5x^2+2x+6)/(x^3+2x^2+x)

Lösung

\int \frac{5 x ^{2} + 2 x + 6}{x ^{3} + 2 x ^{2} + x} d x

- ln ∣ x + 1 ∣ + \frac{9}{x + 1} + 6 ln ∣ x ∣ + C

Schritte zur Lösung

\int \frac{5 x ^{2} + 2 x + 6}{x ^{3} + 2 x ^{2} + x} d x

Multipliziere aus

\frac{5 x ^{2} + 2 x + 6}{x ^{3} + 2 x ^{2} + x} : \frac{5 x ^{2}}{x ^{3} + 2 x ^{2} + x} + \frac{2 x}{x ^{3} + 2 x ^{2} + x} + \frac{6}{x ^{3} + 2 x ^{2} + 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{5 x ^{2}}{x ^{3} + 2 x ^{2} + x} d x + \int \frac{2 x}{x ^{3} + 2 x ^{2} + x} d x + \int \frac{6}{x ^{3} + 2 x ^{2} + x} d x

\int \frac{5 x ^{2}}{x ^{3} + 2 x ^{2} + x} d x = 5 (ln ∣ x + 1 ∣ + \frac{1}{x + 1})

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

\int \frac{6}{x ^{3} + 2 x ^{2} + x} d x = 6 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1})

= 5 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - \frac{2}{x + 1} + 6 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1})

Vereinfache

5 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - \frac{2}{x + 1} + 6 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1}) : - ln ∣ x + 1 ∣ + \frac{9}{x + 1} + 6 ln ∣ x ∣

= - ln ∣ x + 1 ∣ + \frac{9}{x + 1} + 6 ln ∣ x ∣

Füge eine Konstante zur Lösung hinzu

= - ln ∣ x + 1 ∣ + \frac{9}{x + 1} + 6 ln ∣ x ∣ + C

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