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

Lösung

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

\frac{1}{4}

Dezimale

0.25

Schritte zur Lösung

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

Wende U-Substitution an

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

Wende die partielle Integration an

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

\int \frac{1}{2} (u - 2) d u = \frac{u ^{2}}{4} - u

= [ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{1}^{2}

Vereinfache

[ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{1}^{2} : [\frac{1}{2} u^{2} ln (u) - u ln (u) - \frac{u ^{2}}{4} + u]_{1}^{2}

= [\frac{1}{2} u^{2} ln (u) - u ln (u) - \frac{u ^{2}}{4} + u]_{1}^{2}

Berechne die Grenzen:

\frac{1}{4}

= \frac{1}{4}

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