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

Lösung

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

5 ln ∣ x + 1 ∣ - ln x^{2} + 2 x + 2 - 7 arctan (x + 1) + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{3 x ^{2} - x + 1}{( x + 1 ) ( x ^{2} + 2 x + 2 )} : \frac{5}{x + 1} + \frac{- 2 x - 9}{x ^{2} + 2 x + 2}

= \int \frac{5}{x + 1} + \frac{- 2 x - 9}{x ^{2} + 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 \frac{5}{x + 1} d x + \int \frac{- 2 x - 9}{x ^{2} + 2 x + 2} d x

\int \frac{5}{x + 1} d x = 5 ln ∣ x + 1 ∣

\int \frac{- 2 x - 9}{x ^{2} + 2 x + 2} d x = - ln x^{2} + 2 x + 2 - 7 arctan (x + 1)

= 5 ln ∣ x + 1 ∣ - ln x^{2} + 2 x + 2 - 7 arctan (x + 1)

Füge eine Konstante zur Lösung hinzu

= 5 ln ∣ x + 1 ∣ - ln x^{2} + 2 x + 2 - 7 arctan (x + 1) + C