integral of 1/(3x+2)

解答

\int \frac{1}{3 x + 2} d x

\frac{1}{3} ln ∣ 3 x + 2 ∣ + C

求解步骤

\int \frac{1}{3 x + 2} d x

使用换元积分法

= \int \frac{1}{3 u} d u

提出常数:

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

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

使用积分常见定则:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

u = 3 x + 2

代回

= \frac{1}{3} ln ∣ 3 x + 2 ∣

解答补常数

= \frac{1}{3} ln ∣ 3 x + 2 ∣ + C

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