integral of 1/(2x^2+5x-7)

Lösung

\int \frac{1}{2 x ^{2} + 5 x - 7} d x

\frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ 2 x + 7 ∣ + C

Schritte zur Lösung

\int \frac{1}{2 x ^{2} + 5 x - 7} d x

Ermittle den Partialbruch von

\frac{1}{2 x ^{2} + 5 x - 7} : \frac{1}{9 ( x - 1 )} - \frac{2}{9 ( 2 x + 7 )}

= \int \frac{1}{9 ( x - 1 )} - \frac{2}{9 ( 2 x + 7 )} 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}{9 ( x - 1 )} d x - \int \frac{2}{9 ( 2 x + 7 )} d x

\int \frac{1}{9 ( x - 1 )} d x = \frac{1}{9} ln ∣ x - 1 ∣

\int \frac{2}{9 ( 2 x + 7 )} d x = \frac{1}{9} ln ∣ 2 x + 7 ∣

= \frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ 2 x + 7 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{9} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ 2 x + 7 ∣ + C