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

Lösung

integral

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

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

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

Füge eine Konstante zur Lösung hinzu

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

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

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

in t e g r a l \frac{sin ^{3} ( x )}{cos ( x )}

in t e g r a l \frac{x ^{2}}{x ^{2} + 5}

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