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

Lösung

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

ln ∣ x ∣ + 2 ln ∣ x + 3 ∣ - ln ∣ x - 3 ∣ + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{2}{x + 3} d x = 2 ln ∣ x + 3 ∣

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

= ln ∣ x ∣ + 2 ln ∣ x + 3 ∣ - ln ∣ x - 3 ∣

Füge eine Konstante zur Lösung hinzu

= ln ∣ x ∣ + 2 ln ∣ x + 3 ∣ - ln ∣ x - 3 ∣ + C

\int (I n x)^{2} d x

\int cos (x) + (x^{2} + 1)^{7} \cdot 2 x d x

im pl i c i t \frac{d y}{d x}, ln (x y^{3}) = x y

\int \frac{tan ^{3} ( ln ( x ) )}{2 x} d x

d er i v a t i v e f (x) = cos^{5} (x)