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受欢迎的微积分 >

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

解答

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

解答

\frac{1}{2} ln 4 x^{2} - 4 x + 24 + \frac{1}{23} arctan (\frac{2 x - 1}{23}) + C

求解步骤

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

对

x^{2} - x + 6

配方:

(x - \frac{1}{2})^{2} + \frac{23}{4}

= \int \frac{x}{( x - \frac{1}{2} ) ^{2} + \frac{23}{4}} d x

使用换元积分法

= \int \frac{2 ( 2 u + 1 )}{4 u ^{2} + 23} d u

提出常数:

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

= 2 \cdot \int \frac{2 u + 1}{4 u ^{2} + 23} d u

乘开

\frac{2 u + 1}{4 u ^{2} + 23} : \frac{2 u}{4 u ^{2} + 23} + \frac{1}{4 u ^{2} + 23}

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 (\int \frac{2 u}{4 u ^{2} + 23} d u + \int \frac{1}{4 u ^{2} + 23} d u)

\int \frac{2 u}{4 u ^{2} + 23} d u = \frac{1}{4} ln 4 u^{2} + 23

\int \frac{1}{4 u ^{2} + 23} d u = \frac{1}{2 23} arctan (\frac{2}{23} u)

= 2 (\frac{1}{4} ln 4 u^{2} + 23 + \frac{1}{2 23} arctan (\frac{2}{23} u))

u = x - \frac{1}{2}

代回

= 2 (\frac{1}{4} ln 4 (x - \frac{1}{2})^{2} + 23 + \frac{1}{2 23} arctan (\frac{2}{23} (x - \frac{1}{2})))

化简

2 (\frac{1}{4} ln 4 (x - \frac{1}{2})^{2} + 23 + \frac{1}{2 23} arctan (\frac{2}{23} (x - \frac{1}{2}))) : \frac{1}{2} ln 4 x^{2} - 4 x + 24 + \frac{1}{23} arctan (\frac{2 x - 1}{23})

= \frac{1}{2} ln 4 x^{2} - 4 x + 24 + \frac{1}{23} arctan (\frac{2 x - 1}{23})

解答补常数

= \frac{1}{2} ln 4 x^{2} - 4 x + 24 + \frac{1}{23} arctan (\frac{2 x - 1}{23}) + C

作图

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