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受欢迎的微积分 >

integral of 8/(x^{-1)+2}

解答

\int \frac{8}{x ^{- 1} + 2} d x

解答

2 (1 + 2 x - ln ∣ 1 + 2 x ∣) + C

求解步骤

\int \frac{8}{x ^{- 1} + 2} d x

提出常数:

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

= 8 \cdot \int \frac{1}{x ^{- 1} + 2} d x

化简

\frac{1}{x ^{- 1} + 2} : \frac{x}{1 + 2 x}

= 8 \cdot \int \frac{x}{1 + 2 x} d x

使用换元积分法

= 8 \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

= 8 \cdot \frac{1}{4} \cdot \int \frac{u - 1}{u} d u

乘开

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

= 8 \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

= 8 \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 ∣

= 8 \cdot \frac{1}{4} (u - ln ∣ u ∣)

u = 1 + 2 x

代回

= 8 \cdot \frac{1}{4} (1 + 2 x - ln ∣ 1 + 2 x ∣)

化简

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

= 2 (1 + 2 x - ln ∣ 1 + 2 x ∣)

解答补常数

= 2 (1 + 2 x - ln ∣ 1 + 2 x ∣) + C

作图

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