integral of 1/(x^2-x^4)

Lösung

\int \frac{1}{x ^{2} - x ^{4}} d x

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

Schritte zur Lösung

\int \frac{1}{x ^{2} - x ^{4}} d x

Ermittle den Partialbruch von

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

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

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

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

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

= - \frac{1}{x} + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣

Füge eine Konstante zur Lösung hinzu

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

t an g e n t 5 x^{3} - x^{2} y + y^{3} + 1 = 0, (1, - 2)

t an g e n t y = t^{3} - 4 t - 3, (2, - 3)

x \to 2 + lim (\frac{3}{( x - 2 ) ^{3}})

\int (y e^{y}) d y

\int_{0}^{5} π (2 5 y)^{2} d y