integral of (3x+1)/(x+1)

Lösung

\int \frac{3 x + 1}{x + 1} d x

3 x + 3 - 2 ln ∣ x + 1 ∣ + C

Schritte zur Lösung

\int \frac{3 x + 1}{x + 1} d x

Multipliziere aus

\frac{3 x + 1}{x + 1} : \frac{3 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{3 x}{x + 1} d x + \int \frac{1}{x + 1} d x

\int \frac{3 x}{x + 1} d x = 3 (x + 1 - ln ∣ x + 1 ∣)

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

= 3 (x + 1 - ln ∣ x + 1 ∣) + ln ∣ x + 1 ∣

Vereinfache

3 (x + 1 - ln ∣ x + 1 ∣) + ln ∣ x + 1 ∣ : 3 x + 3 - 2 ln ∣ x + 1 ∣

= 3 x + 3 - 2 ln ∣ x + 1 ∣

Füge eine Konstante zur Lösung hinzu

= 3 x + 3 - 2 ln ∣ x + 1 ∣ + C

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