integral from 0 to 1 of (-1+2/(1+t))

解答

\int_{0}^{1} (- 1 + \frac{2}{1 + t}) d t

- 1 + 2 ln (2)

十进制

0.38629 \dots

求解步骤

\int_{0}^{1} - 1 + \frac{2}{1 + t} d t

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}^{1} 1 d t + \int_{0}^{1} \frac{2}{1 + t} d t

\int_{0}^{1} 1 d t = 1

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

= - 1 + 2 ln (2)

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