integral of ((5x-12))/(x^3-6x^2+8x)

Lösung

\int \frac{( 5 x - 12 )}{x ^{3} - 6 x ^{2} + 8 x} d x

\frac{5 x}{x ^{2} - 6 x + 8} + \frac{1}{2} ln ∣ x - 2 ∣ + ln ∣ x - 4 ∣ - \frac{3}{2} ln ∣ x ∣ + \frac{5}{x - 2} - \frac{10}{x - 4} + C

Schritte zur Lösung

\int \frac{5 x - 12}{x ^{3} - 6 x ^{2} + 8 x} d x

Multipliziere aus

\frac{5 x - 12}{x ^{3} - 6 x ^{2} + 8 x} : \frac{5 x}{x ^{3} - 6 x ^{2} + 8 x} - \frac{12}{x ^{3} - 6 x ^{2} + 8 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}{x ^{3} - 6 x ^{2} + 8 x} d x - \int \frac{12}{x ^{3} - 6 x ^{2} + 8 x} d x

\int \frac{5 x}{x ^{3} - 6 x ^{2} + 8 x} d x = 5 (\frac{x}{x ^{2} - 6 x + 8} - \frac{1}{2} ln ∣ x - 2 ∣ + \frac{1}{x - 2} + \frac{1}{2} ln ∣ x - 4 ∣ - \frac{2}{x - 4})

\int \frac{12}{x ^{3} - 6 x ^{2} + 8 x} d x = 12 (\frac{1}{8} ln ∣ x ∣ - \frac{1}{4} ln ∣ x - 2 ∣ + \frac{1}{8} ln ∣ x - 4 ∣)

= 5 (\frac{x}{x ^{2} - 6 x + 8} - \frac{1}{2} ln ∣ x - 2 ∣ + \frac{1}{x - 2} + \frac{1}{2} ln ∣ x - 4 ∣ - \frac{2}{x - 4}) - 12 (\frac{1}{8} ln ∣ x ∣ - \frac{1}{4} ln ∣ x - 2 ∣ + \frac{1}{8} ln ∣ x - 4 ∣)

Vereinfache

5 (\frac{x}{x ^{2} - 6 x + 8} - \frac{1}{2} ln ∣ x - 2 ∣ + \frac{1}{x - 2} + \frac{1}{2} ln ∣ x - 4 ∣ - \frac{2}{x - 4}) - 12 (\frac{1}{8} ln ∣ x ∣ - \frac{1}{4} ln ∣ x - 2 ∣ + \frac{1}{8} ln ∣ x - 4 ∣) : \frac{5 x}{x ^{2} - 6 x + 8} + \frac{1}{2} ln ∣ x - 2 ∣ + ln ∣ x - 4 ∣ - \frac{3}{2} ln ∣ x ∣ + \frac{5}{x - 2} - \frac{10}{x - 4}

= \frac{5 x}{x ^{2} - 6 x + 8} + \frac{1}{2} ln ∣ x - 2 ∣ + ln ∣ x - 4 ∣ - \frac{3}{2} ln ∣ x ∣ + \frac{5}{x - 2} - \frac{10}{x - 4}

Füge eine Konstante zur Lösung hinzu

= \frac{5 x}{x ^{2} - 6 x + 8} + \frac{1}{2} ln ∣ x - 2 ∣ + ln ∣ x - 4 ∣ - \frac{3}{2} ln ∣ x ∣ + \frac{5}{x - 2} - \frac{10}{x - 4} + C

x \to 1 - lim (\frac{3}{x ^{3} - 1})

x y^{^{'}} + y = 6 x y^{2}, y (1) = - 8

(x y^{2} + 3 y^{2}) d y - 2 x d x = 0

s l o p e (4, 1), (7, 10)

\frac{d}{d x} (\frac{18}{x})