integral (x^4)/((x^3+1))

Lösung

integral

\frac{x ^{4}}{( x ^{3} + 1 )}

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

Schritte zur Lösung

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

Lange Division

\frac{x ^{4}}{x ^{3} + 1} : x - \frac{x}{x ^{3} + 1}

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

\int x d x = \frac{x ^{2}}{2}

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

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

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

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

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

in t e g r a l 1 - cot^{2} (x)

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