integral of (4x+1)/(x-1)

Lösung

\int \frac{4 x + 1}{x - 1} d x

4 x - 4 + 5 ln ∣ x - 1 ∣ + C

Schritte zur Lösung

\int \frac{4 x + 1}{x - 1} d x

Multipliziere aus

\frac{4 x + 1}{x - 1} : \frac{4 x}{x - 1} + \frac{1}{x - 1}

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

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

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

= 4 (x - 1 + ln ∣ x - 1 ∣) + ln ∣ x - 1 ∣

Vereinfache

4 (x - 1 + ln ∣ x - 1 ∣) + ln ∣ x - 1 ∣ : 4 x - 4 + 5 ln ∣ x - 1 ∣

= 4 x - 4 + 5 ln ∣ x - 1 ∣

Füge eine Konstante zur Lösung hinzu

= 4 x - 4 + 5 ln ∣ x - 1 ∣ + C

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