integral from-0.5 to 0 of xln(x+1)

Lösung

\int_{- 0.5}^{0} x ln (x + 1) d x

0.05256 \dots

Schritte zur Lösung

\int_{- 0.5}^{0} x ln (x + 1) d x

Wende U-Substitution an

= \int_{0.5}^{1} (u - 1) ln (u) d u

Wende die partielle Integration an

= [ln (u) (0.5 u^{2} - u) - \int 0.5 u - 1 d u]_{0.5}^{1}

\int 0.5 u - 1 d u = 0.25 u^{2} - u

= [ln (u) (0.5 u^{2} - u) - (0.25 u^{2} - u)]_{0.5}^{1}

Multipliziere aus

ln (u) (0.5 u^{2} - u) - (0.25 u^{2} - u) : 0.5 u^{2} ln (u) - u ln (u) - 0.25 u^{2} + u

= [0.5 u^{2} ln (u) - u ln (u) - 0.25 u^{2} + u]_{0.5}^{1}

Berechne die Grenzen:

0.05256 \dots

= 0.05256 \dots

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