integral of (x^2+4)/(3x^3+4x^2-4x)

Lösung

\int \frac{x ^{2} + 4}{3 x ^{3} + 4 x ^{2} - 4 x} d x

- ln ∣ x ∣ + \frac{5}{6} ln ∣ 3 x - 2 ∣ + \frac{1}{2} ln ∣ x + 2 ∣ + C

Schritte zur Lösung

\int \frac{x ^{2} + 4}{3 x ^{3} + 4 x ^{2} - 4 x} d x

Ermittle den Partialbruch von

\frac{x ^{2} + 4}{3 x ^{3} + 4 x ^{2} - 4 x} : - \frac{1}{x} + \frac{5}{2 ( 3 x - 2 )} + \frac{1}{2 ( x + 2 )}

= \int - \frac{1}{x} + \frac{5}{2 ( 3 x - 2 )} + \frac{1}{2 ( x + 2 )} 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{5}{2 ( 3 x - 2 )} d x + \int \frac{1}{2 ( x + 2 )} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{5}{2 ( 3 x - 2 )} d x = \frac{5}{6} ln ∣ 3 x - 2 ∣

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

= - ln ∣ x ∣ + \frac{5}{6} ln ∣ 3 x - 2 ∣ + \frac{1}{2} ln ∣ x + 2 ∣

Füge eine Konstante zur Lösung hinzu

= - ln ∣ x ∣ + \frac{5}{6} ln ∣ 3 x - 2 ∣ + \frac{1}{2} ln ∣ x + 2 ∣ + C