integral of 1/((x^2-4x-5))

Lösung

\int \frac{1}{( x ^{2} - 4 x - 5 )} d x

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

Schritte zur Lösung

\int \frac{1}{x ^{2} - 4 x - 5} d x

Ermittle den Partialbruch von

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

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

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

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

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

Füge eine Konstante zur Lösung hinzu

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

t a y l or x^{4} + 2 x, 1

(12 x + x^{2} + 0.05 x^{3})^{^{'}}

\int_{- 1}^{1} e^{y} - (y^{2} - 2) d y

n = 2 \sum \infty \frac{3}{4 ^{n}}

\frac{\partial}{\partial x} (ln (1 + x^{2} + y^{2}))