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

Lösung

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

ln ∣ x ∣ - ln ∣ x - 1 ∣ + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

= ln ∣ x ∣ - ln ∣ x - 1 ∣

Füge eine Konstante zur Lösung hinzu

= ln ∣ x ∣ - ln ∣ x - 1 ∣ + C

\int \frac{x ^{2} + 58}{x ^{2} + 64} d x

\int - 4 x^{- 2} d x

\int \frac{2 x - 3}{x ^{2} - 3 x + 5} d x

\int \frac{1}{t ^{2} + 81} d t

\int (1 + cos (2 x))^{3} d x