integral from 0 to 1 of x^2ln(x+1)

Lösung

\int_{0}^{1} x^{2} ln (x + 1) d x

- \frac{5}{18} + \frac{2}{3} ln (2)

Dezimale

0.18432 \dots

Schritte zur Lösung

\int_{0}^{1} x^{2} ln (x + 1) d x

Wende U-Substitution an

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

Wende die partielle Integration an

= [ln (u) (\frac{u ^{3}}{3} - u^{2} + u) - \int \frac{1}{3} (u^{2} - 3 u + 3) d u]_{1}^{2}

\int \frac{1}{3} (u^{2} - 3 u + 3) d u = \frac{u ^{3}}{9} - \frac{u ^{2}}{2} + u

= [ln (u) (\frac{u ^{3}}{3} - u^{2} + u) - (\frac{u ^{3}}{9} - \frac{u ^{2}}{2} + u)]_{1}^{2}

Vereinfache

[ln (u) (\frac{u ^{3}}{3} - u^{2} + u) - (\frac{u ^{3}}{9} - \frac{u ^{2}}{2} + u)]_{1}^{2} : [\frac{1}{3} u^{3} ln (u) - u^{2} ln (u) + u ln (u) - \frac{u ^{3}}{9} + \frac{u ^{2}}{2} - u]_{1}^{2}

= [\frac{1}{3} u^{3} ln (u) - u^{2} ln (u) + u ln (u) - \frac{u ^{3}}{9} + \frac{u ^{2}}{2} - u]_{1}^{2}

Berechne die Grenzen:

- \frac{5}{18} + \frac{2}{3} ln (2)

= - \frac{5}{18} + \frac{2}{3} ln (2)

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