integral from 0 to 1 of 5(e^{1/x})/(x^3)

解答

\int_{0}^{1} 5 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

发散

求解步骤

\int_{0}^{1} 5 \cdot \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

提出常数:

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

= 5 \cdot \int_{0}^{1} \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

使用分布积分法

= 5 [\frac{e ^{\frac{1}{x}}}{x ^{2}} - \int - \frac{e ^{\frac{1}{x}} ( 3 x + 1 )}{x ^{4}} d x]_{0}^{1}

\int - \frac{e ^{\frac{1}{x}} ( 3 x + 1 )}{x ^{4}} d x = \frac{3 e ^{\frac{1}{x}}}{x} - e^{\frac{1}{x}} + \frac{e ^{\frac{1}{x}}}{x ^{2}} - \frac{2 e ^{\frac{1}{x}}}{x}

= 5 [\frac{e ^{\frac{1}{x}}}{x ^{2}} - (\frac{3 e ^{\frac{1}{x}}}{x} - e^{\frac{1}{x}} + \frac{e ^{\frac{1}{x}}}{x ^{2}} - \frac{2 e ^{\frac{1}{x}}}{x})]_{0}^{1}

化简

5 [\frac{e ^{\frac{1}{x}}}{x ^{2}} - (\frac{3 e ^{\frac{1}{x}}}{x} - e^{\frac{1}{x}} + \frac{e ^{\frac{1}{x}}}{x ^{2}} - \frac{2 e ^{\frac{1}{x}}}{x})]_{0}^{1} : 5 [\frac{- e ^{\frac{1}{x}} + e ^{\frac{1}{x}} x}{x}]_{0}^{1}

= 5 [\frac{- e ^{\frac{1}{x}} + e ^{\frac{1}{x}} x}{x}]_{0}^{1}

带入上下限计算后得:

发散

= 发散

d er i v a t i v e (\frac{8 x + 5}{8 x - 5})^{\frac{1}{2}}

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\frac{\partial}{\partial x} (sin (8 x - 9 y))

d er i v a t i v e \frac{x}{2} + 3