integral from 1 to e of tln(t)

解答

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

\frac{e ^{2} + 1}{4}

十进制

2.09726 \dots

求解步骤

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

使用分布积分法

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

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

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

2 \cdot \frac{t ^{2}}{4} = \frac{t ^{2}}{2}

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

带入上下限计算后得:

\frac{e ^{2}}{4} + \frac{1}{4}

= \frac{e ^{2}}{4} + \frac{1}{4}

化简

= \frac{e ^{2} + 1}{4}

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