integral from 0 to 1 of ln|x+1|

Lösung

\int_{0}^{1} ln ∣ x + 1 ∣ d x

2 ln (2) - 1

Dezimale

0.38629 \dots

Schritte zur Lösung

\int_{0}^{1} ln ∣ x + 1 ∣ d x

Entferne die Extremwerte

= \int_{0}^{1} ln (x + 1) d x

Wende U-Substitution an

= \int_{1}^{2} ln (u) d u

Wende die partielle Integration an

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

\int 1 d u = u

= [u ln (u) - u]_{1}^{2}

Berechne die Grenzen:

2 ln (2) - 1

= 2 ln (2) - 1

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