integral from 4 to 9 of 1/(sqrt(x)-1)

解答

\int_{4}^{9} \frac{1}{x - 1} d x

2 (1 + ln (2))

十进制

3.38629 \dots

求解步骤

\int_{4}^{9} \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} (x^{3} - 4 x^{2}) d x

\int_{- 1}^{2} (9 x^{2} - 4 x + 5) d x

\int_{- 4}^{4} (- x^{2} + 16) d x

\int_{1}^{5} (5 + \frac{1}{x} + e^{x}) d x

\int_{4.9}^{7.1} (12 x^{2} - 6 x + 4) d x