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

Lösung

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

\frac{1}{3} x + \frac{2}{9} + \frac{1}{9} ln ∣ 3 x + 2 ∣ + C

Schritte zur Lösung

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

Multipliziere aus

\frac{x + 1}{3 x + 2} : \frac{x}{3 x + 2} + \frac{1}{3 x + 2}

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

\int \frac{x}{3 x + 2} d x = \frac{1}{9} (3 x + 2 - 2 ln ∣ 3 x + 2 ∣)

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

= \frac{1}{9} (3 x + 2 - 2 ln ∣ 3 x + 2 ∣) + \frac{1}{3} ln ∣ 3 x + 2 ∣

Vereinfache

\frac{1}{9} (3 x + 2 - 2 ln ∣ 3 x + 2 ∣) + \frac{1}{3} ln ∣ 3 x + 2 ∣ : \frac{1}{3} x + \frac{2}{9} + \frac{1}{9} ln ∣ 3 x + 2 ∣

= \frac{1}{3} x + \frac{2}{9} + \frac{1}{9} ln ∣ 3 x + 2 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} x + \frac{2}{9} + \frac{1}{9} ln ∣ 3 x + 2 ∣ + C

\int_{- 3}^{1} x d x

x \to \infty lim (2 ln (x))

x \to - 6 lim (5 x^{2} + 6 x + a)

\int_{0}^{e} \frac{1}{x + e} d x

\int \frac{x}{1 + 9 x ^{2}} d x