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

Lösung

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

ln ∣ x + 1 ∣

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{1}^{1 + x} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{1 + x}

Berechne die Grenzen:

ln ∣ x + 1 ∣

= ln ∣ x + 1 ∣

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