integral of 1/((x+4)(x-4))

Lösung

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

- \frac{1}{8} ln ∣ x + 4 ∣ + \frac{1}{8} ln ∣ x - 4 ∣ + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{1}{( x + 4 ) ( x - 4 )} : - \frac{1}{8 ( x + 4 )} + \frac{1}{8 ( x - 4 )}

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

\int \frac{1}{8 ( x + 4 )} d x = \frac{1}{8} ln ∣ x + 4 ∣

\int \frac{1}{8 ( x - 4 )} d x = \frac{1}{8} ln ∣ x - 4 ∣

= - \frac{1}{8} ln ∣ x + 4 ∣ + \frac{1}{8} ln ∣ x - 4 ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{8} ln ∣ x + 4 ∣ + \frac{1}{8} ln ∣ x - 4 ∣ + C

d er i v a t i v e y = 2 x arccos (x) - 2 1 - x^{2}

n = 0 \sum \infty \frac{( n )}{n + 1}

\frac{y}{x ^{2}} \frac{d y}{d x} + e^{2 x^{3} + y^{2}} = 0

x \to \infty lim (\frac{4}{x} - 12)

\frac{\partial}{\partial y} (3 x + 2 y sin (x))