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受欢迎的微积分 >

integral of ((10x+2))/((3x+1)(x-1))

解答

\int \frac{( 10 x + 2 )}{( 3 x + 1 ) ( x - 1 )} d x

解答

\frac{10}{3} (ln ∣ 30 x - 30 ∣ - 4 (- \frac{ln ∣ 30 x + 10 ∣}{40} + \frac{ln ∣ 30 x - 30 ∣}{40})) + C

求解步骤

\int \frac{10 x + 2}{( 3 x + 1 ) ( x - 1 )} d x

使用换元积分法

= \int \frac{10 u}{( 3 u + 4 ) ( u - 12 )} d u

提出常数:

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

= 10 \cdot \int \frac{u}{( 3 u + 4 ) ( u - 12 )} d u

使用换元积分法

= 10 \cdot \int \frac{v - 4}{3 v ( v - 40 )} d v

提出常数:

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

= 10 \cdot \frac{1}{3} \cdot \int \frac{v - 4}{v ( v - 40 )} d v

乘开

\frac{v - 4}{v ( v - 40 )} : \frac{v}{v ( v - 40 )} - \frac{4}{v ( v - 40 )}

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

= 10 \cdot \frac{1}{3} (\int \frac{v}{v ( v - 40 )} d v - \int \frac{4}{v ( v - 40 )} d v)

\int \frac{v}{v ( v - 40 )} d v = ln ∣ v - 40 ∣

\int \frac{4}{v ( v - 40 )} d v = 4 (- \frac{1}{40} ln ∣ v ∣ + \frac{1}{40} ln ∣ v - 40 ∣)

= 10 \cdot \frac{1}{3} (ln ∣ v - 40 ∣ - 4 (- \frac{1}{40} ln ∣ v ∣ + \frac{1}{40} ln ∣ v - 40 ∣))

代回

= 10 \cdot \frac{1}{3} (ln ∣ 3 (10 x + 2) + 4 - 40 ∣ - 4 (- \frac{1}{40} ln ∣ 3 (10 x + 2) + 4 ∣ + \frac{1}{40} ln ∣ 3 (10 x + 2) + 4 - 40 ∣))

化简

10 \cdot \frac{1}{3} (ln ∣ 3 (10 x + 2) + 4 - 40 ∣ - 4 (- \frac{1}{40} ln ∣ 3 (10 x + 2) + 4 ∣ + \frac{1}{40} ln ∣ 3 (10 x + 2) + 4 - 40 ∣)) : \frac{10}{3} (ln ∣ 30 x - 30 ∣ - 4 (- \frac{ln ∣ 30 x + 10 ∣}{40} + \frac{ln ∣ 30 x - 30 ∣}{40}))

= \frac{10}{3} (ln ∣ 30 x - 30 ∣ - 4 (- \frac{ln ∣ 30 x + 10 ∣}{40} + \frac{ln ∣ 30 x - 30 ∣}{40}))

解答补常数

= \frac{10}{3} (ln ∣ 30 x - 30 ∣ - 4 (- \frac{ln ∣ 30 x + 10 ∣}{40} + \frac{ln ∣ 30 x - 30 ∣}{40})) + C

作图

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