integral from 1 to 2 of xln(x-1)

Lösung

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

- \frac{5}{4}

Dezimale

- 1.25

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{0}^{1} (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]_{0}^{1}

\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)]_{0}^{1}

Vereinfache

[ln (u) (\frac{u ^{2}}{2} + u) - (\frac{u ^{2}}{4} + u)]_{0}^{1} : [\frac{1}{2} u^{2} ln (u) + u ln (u) - \frac{u ^{2}}{4} - u]_{0}^{1}

= [\frac{1}{2} u^{2} ln (u) + u ln (u) - \frac{u ^{2}}{4} - u]_{0}^{1}

Berechne die Grenzen:

- \frac{5}{4}

= - \frac{5}{4}

\int_{2}^{x^{2}} 1 + x^{2} d x

\int_{0}^{3} x^{2} - 2 x + 3 d x

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\int_{a}^{a + b} \frac{1}{x} d x