integral from 0 to 5 of (2/(x+1))

Lösung

\int_{0}^{5} (\frac{2}{x + 1}) d x

2 ln (6)

Dezimale

3.58351 \dots

Schritte zur Lösung

\int_{0}^{5} \frac{2}{x + 1} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{5} \frac{1}{x + 1} d x

Wende U-Substitution an

= 2 \cdot \int_{1}^{6} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 [ln ∣ u ∣]_{1}^{6}

Berechne die Grenzen:

ln (6)

= 2 ln (6)

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