integral of (9x^3-6x+2)/(x^3-x^2)

Lösung

\int \frac{9 x ^{3} - 6 x + 2}{x ^{3} - x ^{2}} d x

\frac{9 x ^{2}}{x - 1} + 5 ln ∣ x - 1 ∣ - \frac{9}{x - 1} + 4 ln ∣ x ∣ + \frac{2}{x} + C

Schritte zur Lösung

\int \frac{9 x ^{3} - 6 x + 2}{x ^{3} - x ^{2}} d x

Multipliziere aus

\frac{9 x ^{3} - 6 x + 2}{x ^{3} - x ^{2}} : \frac{9 x ^{3}}{x ^{3} - x ^{2}} - \frac{6 x}{x ^{3} - x ^{2}} + \frac{2}{x ^{3} - x ^{2}}

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

\int \frac{9 x ^{3}}{x ^{3} - x ^{2}} d x = 9 (\frac{x ^{2}}{x - 1} + ln ∣ x - 1 ∣ - \frac{1}{x - 1})

\int \frac{6 x}{x ^{3} - x ^{2}} d x = 6 (- ln ∣ x ∣ + ln ∣ x - 1 ∣)

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

= 9 (\frac{x ^{2}}{x - 1} + ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - 6 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) + 2 (- ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣)

Vereinfache

9 (\frac{x ^{2}}{x - 1} + ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - 6 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) + 2 (- ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣) : \frac{9 x ^{2}}{x - 1} + 5 ln ∣ x - 1 ∣ - \frac{9}{x - 1} + 4 ln ∣ x ∣ + \frac{2}{x}

= \frac{9 x ^{2}}{x - 1} + 5 ln ∣ x - 1 ∣ - \frac{9}{x - 1} + 4 ln ∣ x ∣ + \frac{2}{x}

Füge eine Konstante zur Lösung hinzu

= \frac{9 x ^{2}}{x - 1} + 5 ln ∣ x - 1 ∣ - \frac{9}{x - 1} + 4 ln ∣ x ∣ + \frac{2}{x} + C

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

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

\int x^{\frac{1}{2}} + 8 d x

\int (x cos (2 x)) d x

\int (sin (4 x))^{2} d x