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

解答

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

2 (1 - 4 (ln (5) - 2 ln (2)))

十进制

0.21485 \dots

求解步骤

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

使用换元积分法

= \int_{4}^{5} \frac{2 ( u - 4 )}{u} d u

提出常数:

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

= 2 \cdot \int_{4}^{5} \frac{u - 4}{u} d u

乘开

\frac{u - 4}{u} : 1 - \frac{4}{u}

= 2 \cdot \int_{4}^{5} 1 - \frac{4}{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_{4}^{5} 1 d u - \int_{4}^{5} \frac{4}{u} d u)

\int_{4}^{5} 1 d u = 1

\int_{4}^{5} \frac{4}{u} d u = 4 (ln (5) - 2 ln (2))

= 2 (1 - 4 (ln (5) - 2 ln (2)))

\int_{1}^{4} (x^{2} - 9 x) d x

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