integral of 1/(x(x-2)^2)

Lösung

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

\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x - 2 ∣ - \frac{1}{2 ( x - 2 )} + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

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

\int \frac{1}{4 ( x - 2 )} d x = \frac{1}{4} ln ∣ x - 2 ∣

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

= \frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x - 2 ∣ - \frac{1}{2 ( x - 2 )}

Füge eine Konstante zur Lösung hinzu

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

\int_{0}^{\infty} \frac{1}{x ^{2} - 1} d x

t a y l or (1 + x)^{(- 4)}, 0

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

\int_{- 125}^{64} \frac{1}{x ^{\frac{1}{3}}} d x

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