integral from 1 to t of xln(x)

解答

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

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

求解步骤

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

使用分布积分法

= [\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}

带入上下限计算后得:

\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}

化简

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

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