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

解答

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

2 (1 - ln (2))

十进制

0.61370 \dots

求解步骤

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

使用换元积分法

= \int_{1}^{2} \frac{2 ( u - 1 )}{u} d u

提出常数:

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

= 2 \cdot \int_{1}^{2} \frac{u - 1}{u} d u

乘开

\frac{u - 1}{u} : 1 - \frac{1}{u}

= 2 \cdot \int_{1}^{2} 1 - \frac{1}{u} d u

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

= 2 (\int_{1}^{2} 1 d u - \int_{1}^{2} \frac{1}{u} d u)

\int_{1}^{2} 1 d u = 1

\int_{1}^{2} \frac{1}{u} d u = ln (2)

= 2 (1 - ln (2))

\int_{0}^{1} 2 x + 7 d x

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