integral of 1/(x^3+2x^2-3x)

Lösung

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

- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x + 3 ∣ + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{1}{x ^{3} + 2 x ^{2} - 3 x} : - \frac{1}{3 x} + \frac{1}{4 ( x - 1 )} + \frac{1}{12 ( x + 3 )}

= \int - \frac{1}{3 x} + \frac{1}{4 ( x - 1 )} + \frac{1}{12 ( x + 3 )} 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}{3 x} d x + \int \frac{1}{4 ( x - 1 )} d x + \int \frac{1}{12 ( x + 3 )} d x

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

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

\int \frac{1}{12 ( x + 3 )} d x = \frac{1}{12} ln ∣ x + 3 ∣

= - \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x + 3 ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{12} ln ∣ x + 3 ∣ + C

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

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

\int 3 x cos (nπ x) d x

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

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