integral of 1/(x^3-x^2-9x+9)

Lösung

\int \frac{1}{x ^{3} - x ^{2} - 9 x + 9} d x

- \frac{1}{8} ln ∣ x - 1 ∣ + \frac{1}{24} ln ∣ x + 3 ∣ + \frac{1}{12} ln ∣ x - 3 ∣ + C

Schritte zur Lösung

\int \frac{1}{x ^{3} - x ^{2} - 9 x + 9} d x

Ermittle den Partialbruch von

\frac{1}{x ^{3} - x ^{2} - 9 x + 9} : - \frac{1}{8 ( x - 1 )} + \frac{1}{24 ( x + 3 )} + \frac{1}{12 ( x - 3 )}

= \int - \frac{1}{8 ( x - 1 )} + \frac{1}{24 ( x + 3 )} + \frac{1}{12 ( x - 3 )} 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}{8 ( x - 1 )} d x + \int \frac{1}{24 ( x + 3 )} d x + \int \frac{1}{12 ( x - 3 )} d x

\int \frac{1}{8 ( x - 1 )} d x = \frac{1}{8} ln ∣ x - 1 ∣

\int \frac{1}{24 ( x + 3 )} d x = \frac{1}{24} ln ∣ x + 3 ∣

\int \frac{1}{12 ( x - 3 )} d x = \frac{1}{12} ln ∣ x - 3 ∣

= - \frac{1}{8} ln ∣ x - 1 ∣ + \frac{1}{24} ln ∣ x + 3 ∣ + \frac{1}{12} ln ∣ x - 3 ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{8} ln ∣ x - 1 ∣ + \frac{1}{24} ln ∣ x + 3 ∣ + \frac{1}{12} ln ∣ x - 3 ∣ + C

d er i v a t i v e 8 x^{7} (x - 1)^{7} + 7 (x - 1)^{6} x^{8}

\frac{\partial}{\partial x} (7 y^{8} x^{6} - 6 y^{7} x^{5})

\int 2 x (x^{2} + 2)^{3} d x

n = 1 \sum \infty \frac{1}{n ^{2} + 9}

t an g e n t f (x) = \frac{1}{x - 2}, at x = 4