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受欢迎的微积分 >

integral of 1/(x^2-ax)

解答

\int \frac{1}{x ^{2} - a x} d x

解答

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

求解步骤

\int \frac{1}{x ^{2} - a x} d x

分解

\frac{1}{x ^{2} - a x}

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

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

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

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}{x ( \frac{x}{a} - 1 )} d x

使用换元积分法

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

将

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

用部份分式展开:

- \frac{1}{u} + \frac{1}{u - 1}

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

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}{u} d u + \int \frac{1}{u - 1} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

u = \frac{x}{a}

代回

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

使用绝对值运算法则:

\frac{a}{b} = \frac{∣ a ∣}{∣ b ∣}

\frac{x}{a} = \frac{∣ x ∣}{∣ a ∣}

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

解答补常数

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

作图

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