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

Lösung

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

- \frac{t ^{2} - 2 t - 4}{4} + \frac{t ^{2} ln ( t - 1 )}{2} - \frac{1}{2} ln (t - 1) - t + 1

Schritte zur Lösung

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

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Wende U-Substitution an

= - \int_{t - 1}^{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]_{t - 1}^{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)]_{t - 1}^{1}

Vereinfache

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

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

Berechne die Grenzen:

- \frac{5}{4} - \frac{1}{2} t^{2} ln (t - 1) + \frac{1}{2} ln (t - 1) + t + \frac{( t - 1 ) ^{2}}{4} - 1

= - (- \frac{5}{4} - \frac{1}{2} t^{2} ln (t - 1) + \frac{1}{2} ln (t - 1) + t + \frac{( t - 1 ) ^{2}}{4} - 1)

Vereinfache

= - \frac{t ^{2} - 2 t - 4}{4} + \frac{t ^{2} ln ( t - 1 )}{2} - \frac{1}{2} ln (t - 1) - t + 1

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