integral from-0 to e-1 of 1/(t+1)

Lösung

\int_{- 0}^{e - 1} \frac{1}{t + 1} d t

1

Schritte zur Lösung

\int_{- 0}^{e - 1} \frac{1}{t + 1} d t

Wende U-Substitution an

= \int_{1}^{e} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{e}

Berechne die Grenzen:

1

= 1

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