integral 1/((x^2-6x+5))

Lösung

integral

\frac{1}{( x ^{2} - 6 x + 5 )}

- \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ x - 5 ∣ + C

Schritte zur Lösung

\int \frac{1}{x ^{2} - 6 x + 5} d x

Ermittle den Partialbruch von

\frac{1}{x ^{2} - 6 x + 5} : - \frac{1}{4 ( x - 1 )} + \frac{1}{4 ( x - 5 )}

= \int - \frac{1}{4 ( x - 1 )} + \frac{1}{4 ( x - 5 )} 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 - 1 )} d x + \int \frac{1}{4 ( x - 5 )} d x

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

\int \frac{1}{4 ( x - 5 )} d x = \frac{1}{4} ln ∣ x - 5 ∣

= - \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ x - 5 ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} ln ∣ x - 5 ∣ + C

in t e g r a l \frac{5}{x + 1}

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

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

in t e g r a l \frac{x ^{- 2}}{3}

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