integral from 0 to 1 of sqrt(2+1/(4x))

解答

\int_{0}^{1} 2 + \frac{1}{4 x} d x

\frac{1}{2} (\frac{1}{2 2} ln (22 + 3) + 3)

十进制

1.81161 \dots

求解步骤

\int_{0}^{1} 2 + \frac{1}{4 x} d x

化简

2 + \frac{1}{4 x} : \frac{8 x + 1}{2 x}

= \int_{0}^{1} \frac{8 x + 1}{2 x} d x

提出常数:

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

= \frac{1}{2} \cdot \int_{0}^{1} \frac{8 x + 1}{x} d x

使用分布积分法

= \frac{1}{2} [2 x 8 x + 1 - \int \frac{8 x}{8 x + 1} d x]_{0}^{1}

\int \frac{8 x}{8 x + 1} d x = - \frac{1}{2 2} ln 22 x + 1 + 8 x + x 1 + 8 x

= \frac{1}{2} [2 x 8 x + 1 - (- \frac{1}{2 2} ln 22 x + 1 + 8 x + x 1 + 8 x)]_{0}^{1}

化简

\frac{1}{2} [2 x 8 x + 1 - (- \frac{1}{2 2} ln 22 x + 1 + 8 x + x 1 + 8 x)]_{0}^{1} : \frac{1}{2} [x 8 x + 1 + \frac{1}{2 2} ln 22 x + 8 x + 1]_{0}^{1}

= \frac{1}{2} [x 8 x + 1 + \frac{1}{2 2} ln 22 x + 8 x + 1]_{0}^{1}

带入上下限计算后得:

\frac{1}{2 2} ln (22 + 3) + 3

= \frac{1}{2} (\frac{1}{2 2} ln (22 + 3) + 3)

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