integral of (x+1)/(x-5)

Lösung

\int \frac{x + 1}{x - 5} d x

6 ln ∣ x - 5 ∣ + x + C

Schritte zur Lösung

\int \frac{x + 1}{x - 5} d x

\frac{x + 1}{x - 5} = \frac{6}{x - 5} + 1

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

\int \frac{6}{x - 5} d x = 6 ln ∣ x - 5 ∣

\int 1 d x = x

= 6 ln ∣ x - 5 ∣ + x

Füge eine Konstante zur Lösung hinzu

= 6 ln ∣ x - 5 ∣ + x + C

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