integral from 1 to e^x of 8ln(x)

解答

\int_{1}^{e^{x}} 8 ln (x) d x

8 (e^{x} x - e^{x} + 1)

求解步骤

\int_{1}^{e^{x}} 8 ln (x) d x

提出常数:

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

= 8 \cdot \int_{1}^{e^{x}} ln (x) d x

使用分布积分法

= 8 [x ln (x) - \int 1 d x]_{1}^{e^{x}}

\int 1 d x = x

= 8 [x ln (x) - x]_{1}^{e^{x}}

带入上下限计算后得:

e^{x} x - e^{x} + 1

= 8 (e^{x} x - e^{x} + 1)

\int_{1}^{3} \frac{x}{( 3 x ^{2} - 1 ) ^{3}} d x

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