integral of (3x^3)/(x^3-4x)

Lösung

\int \frac{3 x ^{3}}{x ^{3} - 4 x} d x

3 (\frac{x ^{2}}{x + 2} - \frac{4}{x + 2} + ln ∣ x - 2 ∣ - ln ∣ x + 2 ∣) + C

Schritte zur Lösung

\int \frac{3 x ^{3}}{x ^{3} - 4 x} d x

Entferne die Konstante:

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

= 3 \cdot \int \frac{x ^{3}}{x ^{3} - 4 x} d x

Faktorisiere

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

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

Streiche die gemeinsamen Faktoren:

x

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

Wende U-Substitution an

= 3 \cdot \int \frac{( u - 2 ) ^{2}}{u ( u - 4 )} d u

Wende U-Substitution an

= 3 \cdot \int \frac{( v + 2 ) ^{2}}{v ( v + 4 )} d v

Wende die partielle Integration an

= 3 (\frac{( v + 2 ) ^{2}}{v + 4} - \int \frac{- 8 v - 16}{v ( v + 4 ) ^{2}} d v)

\int \frac{- 8 v - 16}{v ( v + 4 ) ^{2}} d v = \frac{4}{v + 4} - ln ∣ v ∣ + ln ∣ v + 4 ∣

= 3 (\frac{( v + 2 ) ^{2}}{v + 4} - (\frac{4}{v + 4} - ln ∣ v ∣ + ln ∣ v + 4 ∣))

Ersetze zurück

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

ma c l a u r in (6 n - 1) (- 2 x)^{n}

\frac{d y}{d x} = k y^{2}

t an g e n t f (x) = 2 x, (8, 4)

\int y^{2} e^{x y} + cos (x) d x

\frac{\partial}{\partial y} (1 - y^{3})