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受欢迎的微积分 >

integral of 1/(f(f-u))

解答

\int \frac{1}{f ( f - u )} df

解答

\frac{1}{u} (- ln (\frac{∣ f ∣}{∣ u ∣}) + ln \frac{f}{u} - 1) + C

求解步骤

\int \frac{1}{f ( f - u )} df

- u + f = u (- 1 + \frac{f}{u})

= \int \frac{1}{u ( - 1 + \frac{f}{u} ) f} df

提出常数:

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

= \frac{1}{u} \cdot \int \frac{1}{( - 1 + \frac{f}{u} ) f} df

使用换元积分法

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

v = \frac{f}{u}

代回

= \frac{1}{u} (- ln \frac{f}{u} + ln \frac{f}{u} - 1)

使用绝对值运算法则:

\frac{a}{b} = \frac{∣ a ∣}{∣ b ∣}

\frac{f}{u} = \frac{∣ f ∣}{∣ u ∣}

= \frac{1}{u} (- ln (\frac{∣ f ∣}{∣ u ∣}) + ln \frac{f}{u} - 1)

解答补常数

= \frac{1}{u} (- ln (\frac{∣ f ∣}{∣ u ∣}) + ln \frac{f}{u} - 1) + C

作图

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