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

Lösung

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

\frac{1}{2} ln - 4 x^{2} + 4 x + 4 + \frac{1}{2 5} ln \frac{1}{5} (2 x - 1) + 1 - \frac{1}{2 5} ln \frac{1}{5} (2 x - 1) - 1 + C

Schritte zur Lösung

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

Multipliziere aus

\frac{x - 1}{x ^{2} - x - 1} : \frac{x}{x ^{2} - x - 1} - \frac{1}{x ^{2} - x - 1}

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

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

\int \frac{1}{x ^{2} - x - 1} d x = - \frac{1}{5} (ln \frac{2 x - 1}{5} + 1 - ln \frac{2 x - 1}{5} - 1)

= - 2 (- \frac{1}{4} ln - 4 x^{2} + 4 x + 4 + \frac{1}{4 5} (ln \frac{2 x - 1}{5} + 1 - ln \frac{2 x - 1}{5} - 1)) - (- \frac{1}{5} (ln \frac{2 x - 1}{5} + 1 - ln \frac{2 x - 1}{5} - 1))

Vereinfache

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

= \frac{1}{2} ln - 4 x^{2} + 4 x + 4 + \frac{1}{2 5} ln \frac{1}{5} (2 x - 1) + 1 - \frac{1}{2 5} ln \frac{1}{5} (2 x - 1) - 1

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln - 4 x^{2} + 4 x + 4 + \frac{1}{2 5} ln \frac{1}{5} (2 x - 1) + 1 - \frac{1}{2 5} ln \frac{1}{5} (2 x - 1) - 1 + C

t an g e n t 3 x^{3} - 6 x^{2} + 4 x + 21

x \to 0 lim (\frac{18}{x ^{2}})

\frac{\partial}{\partial x} (3 x^{2} - 24 y)

\frac{d}{d x} ((x - 1)^{\frac{1}{3}} (x + 2)^{\frac{2}{3}})

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