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

Lösung

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

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

Ermittle den Partialbruch von

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

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

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

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

\int \frac{x ^{2}}{x ^{2} - 7} d x

\int \frac{1}{2} cos^{3} (x) sin (x) d x

\int \frac{4}{1 - 16 x ^{2}} d x

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