integral 1/((x^2-4x))

Lösung

integral

\frac{1}{( x ^{2} - 4 x )}

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

Schritte zur Lösung

\int \frac{1}{x ^{2} - 4 x} d x

Ermittle den Partialbruch von

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

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

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

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

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

Füge eine Konstante zur Lösung hinzu

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

in t e g r a l (2 x + 3)^{4}

in t e g r a l e \cdot x^{2}

in t e g r a l e^{2 x^{2}}

in t e g r a l x^{2} + 6 x

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