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

解

\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

解答ステップ

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

定数を除く:

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

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

u置換積分法を適用する

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

部分積分を適用する

= - [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}

簡素化

- [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}

限度を計算する:

- \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)

簡素化

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