integral from 0 to 8 of 1/(1+sqrt(1+x))

解答

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

- 2 ln (2) + 4

十进制

2.61370 \dots

求解步骤

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

\frac{1}{1 + 1 + x} = - \frac{x + 1}{1 + 1 + x} + 1

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

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int_{0}^{8} 1 d x = 8

= - 2 (ln (2) + 2) + 8

化简

- 2 (ln (2) + 2) + 8 : - 2 ln (2) + 4

= - 2 ln (2) + 4

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