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受欢迎的微积分 >

integral of (2x+18)/(x^2+10x+24)

解答

\int \frac{2 x + 18}{x ^{2} + 10 x + 24} d x

解答

ln x^{2} + 10 x + 24 - 8 (\frac{1}{2} ln ∣ x + 6 ∣ - \frac{1}{2} ln ∣ x + 4 ∣) + C

求解步骤

\int \frac{2 x + 18}{x ^{2} + 10 x + 24} d x

对

x^{2} + 10 x + 24

配方:

(x + 5)^{2} - 1

= \int \frac{2 x + 18}{( x + 5 ) ^{2} - 1} d x

使用换元积分法

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

乘开

\frac{2 u + 8}{u ^{2} - 1} : \frac{2 u}{u ^{2} - 1} + \frac{8}{u ^{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

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

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

\int \frac{8}{u ^{2} - 1} d u = - 8 (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2} ln ∣ u - 1 ∣)

= ln u^{2} - 1 - 8 (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2} ln ∣ u - 1 ∣)

u = x + 5

代回

= ln (x + 5)^{2} - 1 - 8 (\frac{1}{2} ln ∣ x + 5 + 1 ∣ - \frac{1}{2} ln ∣ x + 5 - 1 ∣)

化简

ln (x + 5)^{2} - 1 - 8 (\frac{1}{2} ln ∣ x + 5 + 1 ∣ - \frac{1}{2} ln ∣ x + 5 - 1 ∣) : ln x^{2} + 10 x + 24 - 8 (\frac{1}{2} ln ∣ x + 6 ∣ - \frac{1}{2} ln ∣ x + 4 ∣)

= ln x^{2} + 10 x + 24 - 8 (\frac{1}{2} ln ∣ x + 6 ∣ - \frac{1}{2} ln ∣ x + 4 ∣)

解答补常数

= ln x^{2} + 10 x + 24 - 8 (\frac{1}{2} ln ∣ x + 6 ∣ - \frac{1}{2} ln ∣ x + 4 ∣) + C

作图

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