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

Lösung

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

3 x - ln ∣ x + 1 ∣ - 2 ln ∣ x + 2 ∣ + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{3 x ^{2} + 6 x + 2}{x ^{2} + 3 x + 2} : 3 - \frac{1}{x + 1} - \frac{2}{x + 2}

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

\int 3 d x = 3 x

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

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

= 3 x - ln ∣ x + 1 ∣ - 2 ln ∣ x + 2 ∣

Füge eine Konstante zur Lösung hinzu

= 3 x - ln ∣ x + 1 ∣ - 2 ln ∣ x + 2 ∣ + C

\frac{d}{d x} (\frac{5}{x ^{2} + 1})

\frac{d}{d x} (9 x - 1)

d er i v a t i v e y = 2 (x^{2} - 7 x) (2 x - 7)

\int cos (x) csc^{9} (x) d x

x^{3} y^{^{'''}} + 6 y = 0