integral of (18)/(9x^2-x^4)

Lösung

\int \frac{18}{9 x ^{2} - x ^{4}} d x

18 (- \frac{1}{9 x} + \frac{1}{54} ln ∣ x + 3 ∣ - \frac{1}{54} ln ∣ x - 3 ∣) + C

Schritte zur Lösung

\int \frac{18}{9 x ^{2} - x ^{4}} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 18 \cdot \int \frac{1}{9 x ^{2} - x ^{4}} d x

Ermittle den Partialbruch von

\frac{1}{9 x ^{2} - x ^{4}} : \frac{1}{9 x ^{2}} + \frac{1}{54 ( x + 3 )} - \frac{1}{54 ( x - 3 )}

= 18 \cdot \int \frac{1}{9 x ^{2}} + \frac{1}{54 ( x + 3 )} - \frac{1}{54 ( 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

= 18 (\int \frac{1}{9 x ^{2}} d x + \int \frac{1}{54 ( x + 3 )} d x - \int \frac{1}{54 ( x - 3 )} d x)

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

\int \frac{1}{54 ( x + 3 )} d x = \frac{1}{54} ln ∣ x + 3 ∣

\int \frac{1}{54 ( x - 3 )} d x = \frac{1}{54} ln ∣ x - 3 ∣

= 18 (- \frac{1}{9 x} + \frac{1}{54} ln ∣ x + 3 ∣ - \frac{1}{54} ln ∣ x - 3 ∣)

Füge eine Konstante zur Lösung hinzu

= 18 (- \frac{1}{9 x} + \frac{1}{54} ln ∣ x + 3 ∣ - \frac{1}{54} ln ∣ x - 3 ∣) + C

x \to 0 - lim (e^{\frac{x}{x + 1}})

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

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

\int e^{2 x} 1 + e^{x} d x

x \to 0 lim (\frac{x - sin ( x )}{x})