integral of 1/((x+6)(x+8))

Lösung

\int \frac{1}{( x + 6 ) ( x + 8 )} d x

\frac{1}{2} ln ∣ x + 6 ∣ - \frac{1}{2} ln ∣ x + 8 ∣ + C

Schritte zur Lösung

\int \frac{1}{( x + 6 ) ( x + 8 )} d x

Ermittle den Partialbruch von

\frac{1}{( x + 6 ) ( x + 8 )} : \frac{1}{2 ( x + 6 )} - \frac{1}{2 ( x + 8 )}

= \int \frac{1}{2 ( x + 6 )} - \frac{1}{2 ( x + 8 )} 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}{2 ( x + 6 )} d x - \int \frac{1}{2 ( x + 8 )} d x

\int \frac{1}{2 ( x + 6 )} d x = \frac{1}{2} ln ∣ x + 6 ∣

\int \frac{1}{2 ( x + 8 )} d x = \frac{1}{2} ln ∣ x + 8 ∣

= \frac{1}{2} ln ∣ x + 6 ∣ - \frac{1}{2} ln ∣ x + 8 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln ∣ x + 6 ∣ - \frac{1}{2} ln ∣ x + 8 ∣ + C

\int_{0}^{5} \frac{1}{x ^{\frac{11}{12}}} d x

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

\frac{d}{d x} (ln (x^{13}))

n = 1001 \sum \infty \frac{1}{n}

\int_{0}^{5} x^{3} x^{2} + 25 d x