integral of a/(bx+c)

解答

\int \frac{a}{b x + c} d x

a \frac{1}{b} ln ∣ b x + c ∣ + C

求解步骤

\int \frac{a}{b x + c} d x

提出常数:

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

= a \cdot \int \frac{1}{b x + c} d x

使用换元积分法

= a \cdot \int \frac{1}{b u} d u

提出常数:

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

= a \frac{1}{b} \cdot \int \frac{1}{u} d u

使用积分常见定则:

\int \frac{1}{u} d u = ln (∣ u ∣)

= a \frac{1}{b} ln ∣ u ∣

u = b x + c

代回

= a \frac{1}{b} ln ∣ b x + c ∣

解答补常数

= a \frac{1}{b} ln ∣ b x + c ∣ + C

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