integral of 1/(x^2+7x+6)

Lösung

\int \frac{1}{x ^{2} + 7 x + 6} d x

\frac{1}{5} ln ∣ x + 1 ∣ - \frac{1}{5} ln ∣ x + 6 ∣ + C

Schritte zur Lösung

\int \frac{1}{x ^{2} + 7 x + 6} d x

Ermittle den Partialbruch von

\frac{1}{x ^{2} + 7 x + 6} : \frac{1}{5 ( x + 1 )} - \frac{1}{5 ( x + 6 )}

= \int \frac{1}{5 ( x + 1 )} - \frac{1}{5 ( x + 6 )} 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}{5 ( x + 1 )} d x - \int \frac{1}{5 ( x + 6 )} d x

\int \frac{1}{5 ( x + 1 )} d x = \frac{1}{5} ln ∣ x + 1 ∣

\int \frac{1}{5 ( x + 6 )} d x = \frac{1}{5} ln ∣ x + 6 ∣

= \frac{1}{5} ln ∣ x + 1 ∣ - \frac{1}{5} ln ∣ x + 6 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{5} ln ∣ x + 1 ∣ - \frac{1}{5} ln ∣ x + 6 ∣ + C

x \to - 2 lim (\frac{8}{x - 3})

\frac{\partial}{\partial u} (u^{9} + v^{5})

\frac{d}{d x} (\frac{x + 8}{3 3 x - 4})

\int_{0}^{π} x + x cos (x) d x

x \to 5 lim (\frac{4}{x})