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受欢迎的微积分 >

integral of (4x)/(1-2x)

解答

\int \frac{4 x}{1 - 2 x} d x

解答

1 - 2 x - ln ∣ 1 - 2 x ∣ + C

求解步骤

\int \frac{4 x}{1 - 2 x} d x

提出常数:

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

= 4 \cdot \int \frac{x}{1 - 2 x} d x

使用换元积分法

= 4 \cdot \int \frac{u - 1}{4 u} d u

提出常数:

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

= 4 \cdot \frac{1}{4} \cdot \int \frac{u - 1}{u} d u

乘开

\frac{u - 1}{u} : 1 - \frac{1}{u}

= 4 \cdot \frac{1}{4} \cdot \int 1 - \frac{1}{u} 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

= 4 \cdot \frac{1}{4} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 4 \cdot \frac{1}{4} (u - ln ∣ u ∣)

u = 1 - 2 x

代回

= 4 \cdot \frac{1}{4} (1 - 2 x - ln ∣ 1 - 2 x ∣)

化简

4 \cdot \frac{1}{4} (1 - 2 x - ln ∣ 1 - 2 x ∣) : 1 - 2 x - ln ∣ 1 - 2 x ∣

= 1 - 2 x - ln ∣ 1 - 2 x ∣

解答补常数

= 1 - 2 x - ln ∣ 1 - 2 x ∣ + C

作图

流行的例子

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in v erse l a pl a ce \frac{s + 4}{s ^{2} + 4 s + 8}

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\frac{d y}{d x} = 2 x y, y (0) = 1

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泰勒 1/(1-x),12

t a y l or \frac{1}{1 - x}, 12

integral of (2x^3)/(x^2-4)

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