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Beliebt Rechnen >

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

Lösung

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

Lösung

- 4 (\frac{x ^{4}}{2 x ^{3} + x ^{2} - 2 x - 1} + \frac{1}{12} ln ∣ 2 x + 1 ∣ + \frac{1}{12 ( 2 x + 1 )} - \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{6 ( x - 1 )}) + C

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Wende die partielle Integration an

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

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

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

Graph

Beliebte Beispiele

limit as x approaches 1 of 5x^4-5x^2+10

x \to 1 lim (5 x^{4} - 5 x^{2} + 10)

f^'(x)=f(x)^4*(4x^2+x)

f^{^{'}} (x) = f (x)^{4} \cdot (4 x^{2} + x)

integral of (6x-7)/((2x+1)(4x^2+1))

\int \frac{6 x - 7}{( 2 x + 1 ) ( 4 x ^{2} + 1 )} d x

laplacetransform s^2+4s+13

l a pl a ce t r an s f or m s^{2} + 4 s + 13

y^{^{'}} = 3 x + x y