integral of (3x+1)/(x^2+2x)

Lösung

\int \frac{3 x + 1}{x ^{2} + 2 x} d x

\frac{5}{2} ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x ∣ + C

Schritte zur Lösung

\int \frac{3 x + 1}{x ^{2} + 2 x} d x

Multipliziere aus

\frac{3 x + 1}{x ^{2} + 2 x} : \frac{3 x}{x ^{2} + 2 x} + \frac{1}{x ^{2} + 2 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{3 x}{x ^{2} + 2 x} d x + \int \frac{1}{x ^{2} + 2 x} d x

\int \frac{3 x}{x ^{2} + 2 x} d x = 3 ln ∣ x + 2 ∣

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

= 3 ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x ∣

Addiere gleiche Elemente:

3 ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x + 2 ∣ = \frac{5}{2} ln ∣ x + 2 ∣

= \frac{5}{2} ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x ∣

Füge eine Konstante zur Lösung hinzu

= \frac{5}{2} ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x ∣ + C

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