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

Lösung

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

\frac{1}{3} ln ∣ 2 x - 1 ∣ - \frac{1}{3} ln ∣ x + 1 ∣ + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

\int \frac{2}{3 ( 2 x - 1 )} d x = \frac{1}{3} ln ∣ 2 x - 1 ∣

\int \frac{1}{3 ( x + 1 )} d x = \frac{1}{3} ln ∣ x + 1 ∣

= \frac{1}{3} ln ∣ 2 x - 1 ∣ - \frac{1}{3} ln ∣ x + 1 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} ln ∣ 2 x - 1 ∣ - \frac{1}{3} ln ∣ x + 1 ∣ + C

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