integral of (2x^4)/(x^2-2x)

Lösung

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

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

Schritte zur Lösung

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

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Vereinfache

\frac{x ^{4}}{x ^{2} - 2 x} : \frac{x ^{3}}{x - 2}

= 2 \cdot \int \frac{x ^{3}}{x - 2} d x

Wende U-Substitution an

= 2 \cdot \int \frac{( u + 2 ) ^{3}}{u} d u

Multipliziere aus

\frac{( u + 2 ) ^{3}}{u} : u^{2} + 6 u + 12 + \frac{8}{u}

= 2 \cdot \int u^{2} + 6 u + 12 + \frac{8}{u} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int u^{2} d u + \int 6 u d u + \int 12 d u + \int \frac{8}{u} d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 6 u d u = 3 u^{2}

\int 12 d u = 12 u

\int \frac{8}{u} d u = 8 ln ∣ u ∣

= 2 (\frac{u ^{3}}{3} + 3 u^{2} + 12 u + 8 ln ∣ u ∣)

Setze in

u = x - 2

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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