integral (x^6)/((x^3+1))

Lösung

integral

\frac{x ^{6}}{( x ^{3} + 1 )}

\frac{x ^{4}}{4} - x + \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3})) + C

Schritte zur Lösung

\int \frac{x ^{6}}{x ^{3} + 1} d x

Lange Division

\frac{x ^{6}}{x ^{3} + 1} : x^{3} - 1 + \frac{1}{x ^{3} + 1}

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

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

\int 1 d x = x

\int \frac{1}{x ^{3} + 1} d x = \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3}))

= \frac{x ^{4}}{4} - x + \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3}))

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{4}}{4} - x + \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} - 4 x}{3} + 23 arctan (\frac{2 x - 1}{3})) + C

in t e g r a l x - (\frac{x ^{2}}{2})

in t e g r a l x^{- 1} 1

in t e g r a l \frac{( x + 2 )}{( x ^{2} + x ^{2} )}

in t e g r a l 4 x + 1

in t e g r a l \frac{1}{( 1 + 4 y ^{2} )}