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受欢迎的微积分 >

integral of (24)/((399x-1)(399+1))

解答

\int \frac{24}{( 399 x - 1 ) ( 399 + 1 )} d x

解答

\frac{1}{6650} ln ∣ 399 x - 1 ∣ + C

求解步骤

\int \frac{24}{( 399 x - 1 ) ( 399 + 1 )} d x

提出常数:

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

= \frac{24}{399 + 1} \cdot \int \frac{1}{399 x - 1} d x

\frac{24}{399 + 1} = \frac{3}{50}

= \frac{3}{50} \cdot \int \frac{1}{399 x - 1} d x

使用换元积分法

= \frac{3}{50} \cdot \int \frac{1}{399 u} d u

提出常数:

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

= \frac{3}{50} \cdot \frac{1}{399} \cdot \int \frac{1}{u} d u

使用积分常见定则:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{3}{50} \cdot \frac{1}{399} ln ∣ u ∣

u = 399 x - 1

代回

= \frac{3}{50} \cdot \frac{1}{399} ln ∣ 399 x - 1 ∣

化简

\frac{3}{50} \cdot \frac{1}{399} ln ∣ 399 x - 1 ∣ : \frac{1}{6650} ln ∣ 399 x - 1 ∣

= \frac{1}{6650} ln ∣ 399 x - 1 ∣

解答补常数

= \frac{1}{6650} ln ∣ 399 x - 1 ∣ + C

作图

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