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受欢迎的微积分 >

integral of x/(2x+4)

解答

\int \frac{x}{2 x + 4} d x

解答

\frac{1}{2} (x + 2 - 2 ln (2 ∣ x + 2 ∣)) + C

求解步骤

\int \frac{x}{2 x + 4} d x

使用换元积分法

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

提出常数:

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

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

乘开

\frac{u - 4}{u} : 1 - \frac{4}{u}

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

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

\int 1 d u = u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= \frac{1}{4} (u - 4 ln ∣ u ∣)

u = 2 x + 4

代回

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

化简

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

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

解答补常数

= \frac{1}{2} (x + 2 - 2 ln (2 ∣ x + 2 ∣)) + C

作图

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