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

Lösung

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

- \frac{1}{4} ln ∣ x ∣ - \frac{1}{2 x} + \frac{11}{3} ln ∣ x + 1 ∣ - \frac{29}{12} ln ∣ x - 2 ∣ + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{x ^{3} - 9 x ^{2} - 1}{x ^{2} ( x + 1 ) ( x - 2 )} : - \frac{1}{4 x} + \frac{1}{2 x ^{2}} + \frac{11}{3 ( x + 1 )} - \frac{29}{12 ( x - 2 )}

= \int - \frac{1}{4 x} + \frac{1}{2 x ^{2}} + \frac{11}{3 ( x + 1 )} - \frac{29}{12 ( 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

= - \int \frac{1}{4 x} d x + \int \frac{1}{2 x ^{2}} d x + \int \frac{11}{3 ( x + 1 )} d x - \int \frac{29}{12 ( x - 2 )} d x

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

\int \frac{1}{2 x ^{2}} d x = - \frac{1}{2 x}

\int \frac{11}{3 ( x + 1 )} d x = \frac{11}{3} ln ∣ x + 1 ∣

\int \frac{29}{12 ( x - 2 )} d x = \frac{29}{12} ln ∣ x - 2 ∣

= - \frac{1}{4} ln ∣ x ∣ - \frac{1}{2 x} + \frac{11}{3} ln ∣ x + 1 ∣ - \frac{29}{12} ln ∣ x - 2 ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{4} ln ∣ x ∣ - \frac{1}{2 x} + \frac{11}{3} ln ∣ x + 1 ∣ - \frac{29}{12} ln ∣ x - 2 ∣ + C

\frac{\partial}{\partial x} (e^{\frac{x y}{z ^{2}}})

y^{^{''}} = \frac{2 k}{1 - k}

\frac{d}{d x} (x^{2} ln ∣ x ∣)

d er i v a t i v e f (x) = \frac{x ^{2} + 2 x}{( x + 1 ) ^{2}}

\frac{\partial}{\partial x} (4 e^{yz} cos (x yz))