integral of (x+1)/(2x+1)

Lösung

\int \frac{x + 1}{2 x + 1} d x

\frac{1}{2} x + \frac{1}{4} + \frac{1}{4} ln ∣ 2 x + 1 ∣ + C

Schritte zur Lösung

\int \frac{x + 1}{2 x + 1} d x

Multipliziere aus

\frac{x + 1}{2 x + 1} : \frac{x}{2 x + 1} + \frac{1}{2 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{x}{2 x + 1} d x + \int \frac{1}{2 x + 1} d x

\int \frac{x}{2 x + 1} d x = \frac{1}{4} (2 x + 1 - ln ∣ 2 x + 1 ∣)

\int \frac{1}{2 x + 1} d x = \frac{1}{2} ln ∣ 2 x + 1 ∣

= \frac{1}{4} (2 x + 1 - ln ∣ 2 x + 1 ∣) + \frac{1}{2} ln ∣ 2 x + 1 ∣

Vereinfache

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

= \frac{1}{2} x + \frac{1}{4} + \frac{1}{4} ln ∣ 2 x + 1 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} x + \frac{1}{4} + \frac{1}{4} ln ∣ 2 x + 1 ∣ + C

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