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

Lösung

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

ln (3) - ln (2)

Dezimale

0.40546 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{2}^{3} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{2}^{3}

Berechne die Grenzen:

ln (3) - ln (2)

= ln (3) - ln (2)

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