integral of (x+9)/(2x^3-8x)

Lösung

\int \frac{x + 9}{2 x ^{3} - 8 x} d x

\frac{1}{2} (- \frac{9}{4} ln ∣ x ∣ + \frac{7}{8} ln ∣ x + 2 ∣ + \frac{11}{8} ln ∣ x - 2 ∣) + C

Schritte zur Lösung

\int \frac{x + 9}{2 x ^{3} - 8 x} d x

Faktorisiere

\frac{x + 9}{2 x ^{3} - 8 x}

= \int \frac{x + 9}{2 x ( x + 2 ) ( x - 2 )} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{x + 9}{x ( x + 2 ) ( x - 2 )} d x

Ermittle den Partialbruch von

\frac{x + 9}{x ( x + 2 ) ( x - 2 )} : - \frac{9}{4 x} + \frac{7}{8 ( x + 2 )} + \frac{11}{8 ( x - 2 )}

= \frac{1}{2} \cdot \int - \frac{9}{4 x} + \frac{7}{8 ( x + 2 )} + \frac{11}{8 ( x - 2 )} 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

= \frac{1}{2} (- \int \frac{9}{4 x} d x + \int \frac{7}{8 ( x + 2 )} d x + \int \frac{11}{8 ( x - 2 )} d x)

\int \frac{9}{4 x} d x = \frac{9}{4} ln ∣ x ∣

\int \frac{7}{8 ( x + 2 )} d x = \frac{7}{8} ln ∣ x + 2 ∣

\int \frac{11}{8 ( x - 2 )} d x = \frac{11}{8} ln ∣ x - 2 ∣

= \frac{1}{2} (- \frac{9}{4} ln ∣ x ∣ + \frac{7}{8} ln ∣ x + 2 ∣ + \frac{11}{8} ln ∣ x - 2 ∣)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} (- \frac{9}{4} ln ∣ x ∣ + \frac{7}{8} ln ∣ x + 2 ∣ + \frac{11}{8} ln ∣ x - 2 ∣) + C

\int \frac{- x}{3} d x

\int_{6}^{x} \frac{1}{20} + \frac{x}{40} d x

\frac{\partial}{\partial x} (\frac{3}{x ^{2}})

\int sin^{8} (x) cos (x) d x

\frac{\partial}{\partial x} (2 arctan (x))