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受欢迎的微积分 >

integral of (-x+2)/(x^2-x+1)

解答

\int \frac{- x + 2}{x ^{2} - x + 1} d x

解答

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

求解步骤

\int \frac{- x + 2}{x ^{2} - x + 1} d x

对

x^{2} - x + 1

配方:

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

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

使用换元积分法

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

使用换元积分法

= 2 \cdot \int \frac{- v + 3}{2 ( v ^{2} + 1 )} d v

提出常数:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{- v + 3}{v ^{2} + 1} d v

乘开

\frac{- v + 3}{v ^{2} + 1} : - \frac{v}{v ^{2} + 1} + \frac{3}{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

= 2 \cdot \frac{1}{2} (- \int \frac{v}{v ^{2} + 1} d v + \int \frac{3}{v ^{2} + 1} d v)

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

\int \frac{3}{v ^{2} + 1} d v = 3 arctan (v)

= 2 \cdot \frac{1}{2} (- \frac{1}{2} ln v^{2} + 1 + 3 arctan (v))

代回

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

化简

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

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

解答补常数

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

作图

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