integral from 1 to e-1 of ln(x+1)

Lösung

\int_{1}^{e - 1} ln (x + 1) d x

- 2 ln (2) + 2

Dezimale

0.61370 \dots

Schritte zur Lösung

\int_{1}^{e - 1} ln (x + 1) d x

Wende U-Substitution an

= \int_{2}^{e} ln (u) d u

Wende die partielle Integration an

= [u ln (u) - \int 1 d u]_{2}^{e}

\int 1 d u = u

= [u ln (u) - u]_{2}^{e}

Berechne die Grenzen:

- (2 ln (2) - 2)

= - (2 ln (2) - 2)

Vereinfache

= - 2 ln (2) + 2

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