integral of (x^3-4x+1)/(x^2-3x+2)

Lösung

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

\frac{x ^{2}}{2} + 3 x - ln ∣ x - 1 ∣ + 8 ln ∣ x - 2 ∣ - 2 ln 4 x^{2} - 12 x + 8 + 5 ln ∣ 2 x - 2 ∣ - 5 ln ∣ 2 x - 4 ∣ + C

Schritte zur Lösung

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

Multipliziere aus

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

\int \frac{x ^{3}}{x ^{2} - 3 x + 2} d x = \frac{x ^{2}}{2} + 3 x - ln ∣ x - 1 ∣ + 8 ln ∣ x - 2 ∣

\int \frac{4 x}{x ^{2} - 3 x + 2} d x = 8 (\frac{1}{4} ln 4 x^{2} - 12 x + 8 - \frac{3}{4} (ln ∣ 2 x - 2 ∣ - ln ∣ 2 x - 4 ∣))

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

= \frac{x ^{2}}{2} + 3 x - ln ∣ x - 1 ∣ + 8 ln ∣ x - 2 ∣ - 8 (\frac{1}{4} ln 4 x^{2} - 12 x + 8 - \frac{3}{4} (ln ∣ 2 x - 2 ∣ - ln ∣ 2 x - 4 ∣)) - ln ∣ 2 x - 2 ∣ + ln ∣ 2 x - 4 ∣

Vereinfache

\frac{x ^{2}}{2} + 3 x - ln ∣ x - 1 ∣ + 8 ln ∣ x - 2 ∣ - 8 (\frac{1}{4} ln 4 x^{2} - 12 x + 8 - \frac{3}{4} (ln ∣ 2 x - 2 ∣ - ln ∣ 2 x - 4 ∣)) - ln ∣ 2 x - 2 ∣ + ln ∣ 2 x - 4 ∣ : \frac{x ^{2}}{2} + 3 x - ln ∣ x - 1 ∣ + 8 ln ∣ x - 2 ∣ - 2 ln 4 x^{2} - 12 x + 8 + 5 ln ∣ 2 x - 2 ∣ - 5 ln ∣ 2 x - 4 ∣

= \frac{x ^{2}}{2} + 3 x - ln ∣ x - 1 ∣ + 8 ln ∣ x - 2 ∣ - 2 ln 4 x^{2} - 12 x + 8 + 5 ln ∣ 2 x - 2 ∣ - 5 ln ∣ 2 x - 4 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{2}}{2} + 3 x - ln ∣ x - 1 ∣ + 8 ln ∣ x - 2 ∣ - 2 ln 4 x^{2} - 12 x + 8 + 5 ln ∣ 2 x - 2 ∣ - 5 ln ∣ 2 x - 4 ∣ + C

\int t^{2} + 1 d t

\frac{d x}{d t} = x - x^{3}

\int_{0}^{\infty} e^{- 2 a x} d x

x \to 9 + lim (\frac{x + 8}{x + 9})

t an g e n t f (x) = \frac{x - 7}{2 x - 3}, at x = 1