integral from 0 to 1 of 3ln(4x)

解答

\int_{0}^{1} 3 ln (4 x) d x

3 (2 ln (2) - 1)

十进制

1.15888 \dots

求解步骤

\int_{0}^{1} 3 ln (4 x) d x

提出常数:

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

= 3 \cdot \int_{0}^{1} ln (4 x) d x

使用分布积分法

= 3 [x ln (4 x) - \int 1 d x]_{0}^{1}

\int 1 d x = x

= 3 [x ln (4 x) - x]_{0}^{1}

带入上下限计算后得:

2 ln (2) - 1

= 3 (2 ln (2) - 1)

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