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

Lösung

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{x ^{3} - 2 x}{x ^{2} + 3 x + 2} : x - 3 + \frac{1}{x + 1} + \frac{4}{x + 2}

= \int x - 3 + \frac{1}{x + 1} + \frac{4}{x + 2} 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 x d x - \int 3 d x + \int \frac{1}{x + 1} d x + \int \frac{4}{x + 2} d x

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

\int 3 d x = 3 x

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

\int \frac{4}{x + 2} d x = 4 ln ∣ x + 2 ∣

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{1}{x ^{2} 9 x ^{2} - 1} d x

\int tan^{3} (11 π x) d x

\frac{d}{d x} (\frac{1}{4} sin (2 x))

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

y^{^{''}} + y = cot^{2} (x)