integral from 1 to t of xln(x)

解

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

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

解答ステップ

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

部分積分を適用する

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

\int \frac{x}{2} d x = \frac{x ^{2}}{4}

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

限度を計算する:

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

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

簡素化

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