integral of (x^2)/(x^4-6x^2-27)

Lösung

\int \frac{x ^{2}}{x ^{4} - 6 x ^{2} - 27} d x

\frac{1}{4 3} arctan (\frac{x}{3}) - \frac{1}{8} ln ∣ x + 3 ∣ + \frac{1}{8} ln ∣ x - 3 ∣ + C

Schritte zur Lösung

\int \frac{x ^{2}}{x ^{4} - 6 x ^{2} - 27} d x

Ermittle den Partialbruch von

\frac{x ^{2}}{x ^{4} - 6 x ^{2} - 27} : \frac{1}{4 ( x ^{2} + 3 )} - \frac{1}{8 ( x + 3 )} + \frac{1}{8 ( x - 3 )}

= \int \frac{1}{4 ( x ^{2} + 3 )} - \frac{1}{8 ( x + 3 )} + \frac{1}{8 ( 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}{4 ( x ^{2} + 3 )} d x - \int \frac{1}{8 ( x + 3 )} d x + \int \frac{1}{8 ( x - 3 )} d x

\int \frac{1}{4 ( x ^{2} + 3 )} d x = \frac{1}{4 3} arctan (\frac{x}{3})

\int \frac{1}{8 ( x + 3 )} d x = \frac{1}{8} ln ∣ x + 3 ∣

\int \frac{1}{8 ( x - 3 )} d x = \frac{1}{8} ln ∣ x - 3 ∣

= \frac{1}{4 3} arctan (\frac{x}{3}) - \frac{1}{8} ln ∣ x + 3 ∣ + \frac{1}{8} ln ∣ x - 3 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4 3} arctan (\frac{x}{3}) - \frac{1}{8} ln ∣ x + 3 ∣ + \frac{1}{8} ln ∣ x - 3 ∣ + C

y y^{^{'}} + 36 x = 0

\int \frac{20 x ^{3} - 40 x ^{2} + 9}{x ^{2} - 2 x} d x

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

y^{3} - (6 x + 2) + 3 x y^{2} y^{^{'}} = 0

t an g e n t (x^{3} - 4 x)^{8}