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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

\int \frac{5 x ^{2}}{x ^{3} - 2 x ^{2}} d x = 5 ln ∣ x - 2 ∣

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

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

= 5 ln ∣ x - 2 ∣ - 3 (- \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x - 2 ∣) - \frac{1}{2} ln ∣ x ∣ + \frac{1}{x} + \frac{1}{2} ln ∣ x - 2 ∣

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

y^{^{'}} = \frac{y + x ^{2} - y ^{2}}{x}

\frac{d}{d x} (a cos (x))

\frac{d}{d x} (\frac{x - 1}{x} + ln (x))

d y = x y d x

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