integral of (x^5+x+1)/(x^3-8)

Lösung

\int \frac{x ^{5} + x + 1}{x ^{3} - 8} d x

\frac{x ^{3}}{3} + \frac{35}{12} ln ∣ x - 2 ∣ + \frac{1}{12} (\frac{61}{2} ln \frac{1}{3} (x + 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x + 1))) + C

Schritte zur Lösung

\int \frac{x ^{5} + x + 1}{x ^{3} - 8} d x

Ermittle den Partialbruch von

\frac{x ^{5} + x + 1}{x ^{3} - 8} : x^{2} + \frac{35}{12 ( x - 2 )} + \frac{61 x + 64}{12 ( x ^{2} + 2 x + 4 )}

= \int x^{2} + \frac{35}{12 ( x - 2 )} + \frac{61 x + 64}{12 ( x ^{2} + 2 x + 4 )} 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^{2} d x + \int \frac{35}{12 ( x - 2 )} d x + \int \frac{61 x + 64}{12 ( x ^{2} + 2 x + 4 )} d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int \frac{35}{12 ( x - 2 )} d x = \frac{35}{12} ln ∣ x - 2 ∣

\int \frac{61 x + 64}{12 ( x ^{2} + 2 x + 4 )} d x = \frac{1}{12} (\frac{61}{2} ln \frac{1}{3} (x + 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x + 1)))

= \frac{x ^{3}}{3} + \frac{35}{12} ln ∣ x - 2 ∣ + \frac{1}{12} (\frac{61}{2} ln \frac{1}{3} (x + 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x + 1)))

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{3}}{3} + \frac{35}{12} ln ∣ x - 2 ∣ + \frac{1}{12} (\frac{61}{2} ln \frac{1}{3} (x + 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x + 1))) + C

d er i v a t i v e y = (x^{2} + 3) ((2 x + 5)^{6})

\int_{0}^{3} x^{2} - 4 d x

\int 29 x^{28} d x

x \to \frac{1}{3} lim (5 x - 7)

t a y l or (2 - x)^{- 1}