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受欢迎的微积分 >

integral of 1/(x(a+bx))

解答

\int \frac{1}{x ( a + b x )} d x

解答

\frac{1}{a} (- ln (\frac{∣ a + b x ∣}{∣ a ∣}) + ln \frac{a + b x}{a} - 1) + C

求解步骤

\int \frac{1}{x ( a + b x )} d x

使用换元积分法

= \int \frac{1}{u ( u - a )} d u

u - a = a (\frac{u}{a} - 1)

= \int \frac{1}{u a ( \frac{u}{a} - 1 )} d u

提出常数:

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

= \frac{1}{a} \cdot \int \frac{1}{u ( \frac{u}{a} - 1 )} d u

使用换元积分法

= \frac{1}{a} \cdot \int \frac{1}{v ( v - 1 )} d v

将

\frac{1}{v ( v - 1 )}

用部份分式展开:

- \frac{1}{v} + \frac{1}{v - 1}

= \frac{1}{a} \cdot \int - \frac{1}{v} + \frac{1}{v - 1} d v

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

= \frac{1}{a} (- \int \frac{1}{v} d v + \int \frac{1}{v - 1} d v)

\int \frac{1}{v} d v = ln ∣ v ∣

\int \frac{1}{v - 1} d v = ln ∣ v - 1 ∣

= \frac{1}{a} (- ln ∣ v ∣ + ln ∣ v - 1 ∣)

代回

= \frac{1}{a} (- ln \frac{a + b x}{a} + ln \frac{a + b x}{a} - 1)

使用绝对值运算法则:

\frac{a}{b} = \frac{∣ a ∣}{∣ b ∣}

\frac{a + b x}{a} = \frac{∣ a + b x ∣}{∣ a ∣}

= \frac{1}{a} (- ln (\frac{∣ a + b x ∣}{∣ a ∣}) + ln \frac{a + b x}{a} - 1)

解答补常数

= \frac{1}{a} (- ln (\frac{∣ a + b x ∣}{∣ a ∣}) + ln \frac{a + b x}{a} - 1) + C

作图

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