integral of 1/(x^2+6x+5)

Lösung

\int \frac{1}{x ^{2} + 6 x + 5} d x

\frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x + 5 ∣ + C

Schritte zur Lösung

\int \frac{1}{x ^{2} + 6 x + 5} d x

Ermittle den Partialbruch von

\frac{1}{x ^{2} + 6 x + 5} : \frac{1}{4 ( x + 1 )} - \frac{1}{4 ( x + 5 )}

= \int \frac{1}{4 ( x + 1 )} - \frac{1}{4 ( x + 5 )} 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}{4 ( x + 1 )} d x - \int \frac{1}{4 ( x + 5 )} d x

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

\int \frac{1}{4 ( x + 5 )} d x = \frac{1}{4} ln ∣ x + 5 ∣

= \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x + 5 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x + 5 ∣ + C

x \to 4 + lim (\frac{1}{x ^{2} - 16})

\frac{d ^{2}}{d x ^{2}} (\frac{x}{x ^{2} + 4})

\int \frac{64 x ^{2} - 49}{x ^{3}} d x

(1 + y) d x + (1 - x) d y = 0

y^{^{'}} = y ln (\frac{1}{y})