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受欢迎的微积分 >

integral of (x+3)/(x^2-6x+5)

解答

\int \frac{x + 3}{x ^{2} - 6 x + 5} d x

解答

\frac{1}{2} ln x^{2} - 6 x + 5 - 3 (\frac{1}{2} ln \frac{1}{2} (x - 3) + 1 - \frac{1}{2} ln \frac{1}{2} (x - 3) - 1) + C

求解步骤

\int \frac{x + 3}{x ^{2} - 6 x + 5} d x

对

x^{2} - 6 x + 5

配方:

(x - 3)^{2} - 4

= \int \frac{x + 3}{( x - 3 ) ^{2} - 4} d x

使用换元积分法

= \int \frac{u + 6}{u ^{2} - 4} d u

乘开

\frac{u + 6}{u ^{2} - 4} : \frac{u}{u ^{2} - 4} + \frac{6}{u ^{2} - 4}

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{u}{u ^{2} - 4} d u + \int \frac{6}{u ^{2} - 4} d u

\int \frac{u}{u ^{2} - 4} d u = \frac{1}{2} ln u^{2} - 4

\int \frac{6}{u ^{2} - 4} d u = - 3 (\frac{1}{2} ln \frac{u}{2} + 1 - \frac{1}{2} ln \frac{u}{2} - 1)

= \frac{1}{2} ln u^{2} - 4 - 3 (\frac{1}{2} ln \frac{u}{2} + 1 - \frac{1}{2} ln \frac{u}{2} - 1)

u = x - 3

代回

= \frac{1}{2} ln (x - 3)^{2} - 4 - 3 (\frac{1}{2} ln \frac{x - 3}{2} + 1 - \frac{1}{2} ln \frac{x - 3}{2} - 1)

化简

\frac{1}{2} ln (x - 3)^{2} - 4 - 3 (\frac{1}{2} ln \frac{x - 3}{2} + 1 - \frac{1}{2} ln \frac{x - 3}{2} - 1) : \frac{1}{2} ln x^{2} - 6 x + 5 - 3 (\frac{1}{2} ln \frac{1}{2} (x - 3) + 1 - \frac{1}{2} ln \frac{1}{2} (x - 3) - 1)

= \frac{1}{2} ln x^{2} - 6 x + 5 - 3 (\frac{1}{2} ln \frac{1}{2} (x - 3) + 1 - \frac{1}{2} ln \frac{1}{2} (x - 3) - 1)

解答补常数

= \frac{1}{2} ln x^{2} - 6 x + 5 - 3 (\frac{1}{2} ln \frac{1}{2} (x - 3) + 1 - \frac{1}{2} ln \frac{1}{2} (x - 3) - 1) + C

作图

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