integral ((6x-4))/((x^3-4x))

Lösung

integral

\frac{( 6 x - 4 )}{( x ^{3} - 4 x )}

ln ∣ x ∣ - 2 ln ∣ x + 2 ∣ + ln ∣ x - 2 ∣ + C

Schritte zur Lösung

\int \frac{6 x - 4}{x ^{3} - 4 x} d x

Ermittle den Partialbruch von

\frac{6 x - 4}{x ^{3} - 4 x} : \frac{1}{x} - \frac{2}{x + 2} + \frac{1}{x - 2}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

= ln ∣ x ∣ - 2 ln ∣ x + 2 ∣ + ln ∣ x - 2 ∣

Füge eine Konstante zur Lösung hinzu

= ln ∣ x ∣ - 2 ln ∣ x + 2 ∣ + ln ∣ x - 2 ∣ + C

in t e g r a l \frac{( x ^{2} + x )}{( x ^{2} + 1 ) ^{2}}

in t e g r a l \frac{c}{r ^{3}}

in t e g r a l (x - 5) ln (x)

in t e g r a l x^{5} - x + 1

in t e g r a l e^{x} \cdot (\frac{1}{x})