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受欢迎的微积分 >

integral of ((12x+30))/((x^2+2x-8))

解答

\int \frac{( 12 x + 30 )}{( x ^{2} + 2 x - 8 )} d x

解答

6 (ln \frac{1}{9} (x + 1)^{2} - 1 - \frac{1}{2} ln \frac{1}{3} (x + 1) + 1 + \frac{1}{2} ln \frac{1}{3} (x + 1) - 1) + C

求解步骤

\int \frac{12 x + 30}{x ^{2} + 2 x - 8} d x

对

x^{2} + 2 x - 8

配方:

(x + 1)^{2} - 9

= \int \frac{12 x + 30}{( x + 1 ) ^{2} - 9} d x

使用换元积分法

= \int \frac{12 u + 18}{u ^{2} - 9} d u

使用换元积分法

= \int \frac{6 ( 2 v + 1 )}{v ^{2} - 1} d v

提出常数:

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

= 6 \cdot \int \frac{2 v + 1}{v ^{2} - 1} d v

乘开

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

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

= 6 (\int \frac{2 v}{v ^{2} - 1} d v + \int \frac{1}{v ^{2} - 1} d v)

\int \frac{2 v}{v ^{2} - 1} d v = ln v^{2} - 1

\int \frac{1}{v ^{2} - 1} d v = - \frac{1}{2} ln ∣ v + 1 ∣ + \frac{1}{2} ln ∣ v - 1 ∣

= 6 (ln v^{2} - 1 - \frac{1}{2} ln ∣ v + 1 ∣ + \frac{1}{2} ln ∣ v - 1 ∣)

代回

= 6 (ln (\frac{x + 1}{3})^{2} - 1 - \frac{1}{2} ln \frac{x + 1}{3} + 1 + \frac{1}{2} ln \frac{x + 1}{3} - 1)

化简

6 (ln (\frac{x + 1}{3})^{2} - 1 - \frac{1}{2} ln \frac{x + 1}{3} + 1 + \frac{1}{2} ln \frac{x + 1}{3} - 1) : 6 (ln \frac{1}{9} (x + 1)^{2} - 1 - \frac{1}{2} ln \frac{1}{3} (x + 1) + 1 + \frac{1}{2} ln \frac{1}{3} (x + 1) - 1)

= 6 (ln \frac{1}{9} (x + 1)^{2} - 1 - \frac{1}{2} ln \frac{1}{3} (x + 1) + 1 + \frac{1}{2} ln \frac{1}{3} (x + 1) - 1)

解答补常数

= 6 (ln \frac{1}{9} (x + 1)^{2} - 1 - \frac{1}{2} ln \frac{1}{3} (x + 1) + 1 + \frac{1}{2} ln \frac{1}{3} (x + 1) - 1) + C

作图

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