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

Lösung

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

- 1

Schritte zur Lösung

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

Wende U-Substitution an

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

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- e}^{- 1}

Berechne die Grenzen:

- 1

= - 1

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