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受欢迎的微积分 >

integral of (2x)/(5x+1)

解答

\int \frac{2 x}{5 x + 1} d x

解答

\frac{2}{25} (5 x + 1 - ln ∣ 5 x + 1 ∣) + C

求解步骤

\int \frac{2 x}{5 x + 1} d x

提出常数:

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

= 2 \cdot \int \frac{x}{5 x + 1} d x

使用换元积分法

= 2 \cdot \int \frac{u - 1}{25 u} d u

提出常数:

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

= 2 \cdot \frac{1}{25} \cdot \int \frac{u - 1}{u} d u

乘开

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

= 2 \cdot \frac{1}{25} \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

= 2 \cdot \frac{1}{25} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 2 \cdot \frac{1}{25} (u - ln ∣ u ∣)

u = 5 x + 1

代回

= 2 \cdot \frac{1}{25} (5 x + 1 - ln ∣ 5 x + 1 ∣)

化简

2 \cdot \frac{1}{25} (5 x + 1 - ln ∣ 5 x + 1 ∣) : \frac{2}{25} (5 x + 1 - ln ∣ 5 x + 1 ∣)

= \frac{2}{25} (5 x + 1 - ln ∣ 5 x + 1 ∣)

解答补常数

= \frac{2}{25} (5 x + 1 - ln ∣ 5 x + 1 ∣) + C

作图

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