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受欢迎的微积分 >

integral of 1/(x-3x^2)

解答

\int \frac{1}{x - 3 x ^{2}} d x

解答

- ln ∣ 3 x - 1 ∣ + ln (3 ∣ x ∣) + C

求解步骤

\int \frac{1}{x - 3 x ^{2}} d x

因式分解出通项

- x : - x (3 x - 1)

= \int \frac{1}{- x ( 3 x - 1 )} d x

提出常数:

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

= - \int \frac{1}{x ( 3 x - 1 )} d x

使用换元积分法

= - \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}

= - \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

= - (\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 ∣

= - (ln ∣ u ∣ - ln ∣ u + 1 ∣)

u = 3 x - 1

代回

= - (ln ∣ 3 x - 1 ∣ - ln ∣ 3 x - 1 + 1 ∣)

化简

- (ln ∣ 3 x - 1 ∣ - ln ∣ 3 x - 1 + 1 ∣) : - ln ∣ 3 x - 1 ∣ + ln (3 ∣ x ∣)

= - ln ∣ 3 x - 1 ∣ + ln (3 ∣ x ∣)

解答补常数

= - ln ∣ 3 x - 1 ∣ + ln (3 ∣ x ∣) + C

作图

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