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

Lösung

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

\int \frac{2 x}{x ^{2} + x + 1} d x = ln 4 x^{2} + 4 x + 4 - \frac{2}{3} arctan (\frac{2 x + 1}{3})

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

Füge eine Konstante zur Lösung hinzu

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

s l o p e (1, 2), (5, - 3)

t an g e n t y = \frac{1}{x + 1}

\int \frac{7 x ^{2}}{x ^{2} - 4} d x

\int x^{2} + x y^{2} + z d x

d er i v a t i v e \frac{4}{x - 5}