integral of 1/(8+x^3)

Lösung

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

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

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

= \int \frac{1}{12 ( x + 2 )} + \frac{- x + 4}{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 \frac{1}{12 ( x + 2 )} d x + \int \frac{- x + 4}{12 ( x ^{2} - 2 x + 4 )} d x

\int \frac{1}{12 ( x + 2 )} d x = \frac{1}{12} ln ∣ x + 2 ∣

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

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{1}{x \cdot 9 - 4 x ^{2}} d x

(sin (5 t))^{^{'}}

\int \frac{1}{36 e ^{- 3 x} + e ^{3 x}} d x

\frac{d}{d x} (tan (x) - sin (x))

d er i v a t i v e h (x) = - 6 x^{- 2}