integral of (x^2+8x+9)/((x-1)(x+2)^2)

Lösung

\int \frac{x ^{2} + 8 x + 9}{( x - 1 ) ( x + 2 ) ^{2}} d x

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

Schritte zur Lösung

\int \frac{x ^{2} + 8 x + 9}{( x - 1 ) ( x + 2 ) ^{2}} d x

Ermittle den Partialbruch von

\frac{x ^{2} + 8 x + 9}{( x - 1 ) ( x + 2 ) ^{2}} : \frac{2}{x - 1} - \frac{1}{x + 2} + \frac{1}{( x + 2 ) ^{2}}

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

\int \frac{2}{x - 1} d x = 2 ln ∣ x - 1 ∣

\int \frac{1}{x + 2} d x = ln ∣ x + 2 ∣

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

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{x + 8}{x ^{2} + x - 6} d x

\int \frac{- 4 x + 8}{( x ^{2} + 1 ) ( x - 1 )} d x

\int x^{2} - 6 d x

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

\int \frac{x ^{2} + 7 x - 6}{( x + 1 ) ( x ^{2} - 4 x + 4 )} d x