integral 1/(1+x^3)

Lösung

integral

\frac{1}{1 + x ^{3}}

\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{1}{1 + x ^{3}} d x

Ermittle den Partialbruch von

\frac{1}{1 + x ^{3}} : \frac{1}{3 ( x + 1 )} + \frac{- x + 2}{3 ( x ^{2} - x + 1 )}

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

\int \frac{1}{3 ( x + 1 )} d x = \frac{1}{3} ln ∣ x + 1 ∣

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

= \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{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 \frac{( 1 + x )}{( 1 - x )}

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

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

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

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