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受欢迎的微积分 >

integral of ((6-x))/(sqrt(4x^2-12x+7))

解答

\int \frac{( 6 - x )}{4 x ^{2} - 12 x + 7} d x

解答

\frac{1}{4} (- 4 x^{2} - 12 x + 7 + 9 ln 2 (x^{2} - 3 x + \frac{9}{4}) - 1 + 2 (x - \frac{3}{2})) + C

求解步骤

\int \frac{6 - x}{4 x ^{2} - 12 x + 7} d x

对

4 x^{2} - 12 x + 7

配方:

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

= \int \frac{6 - x}{4 ( x - \frac{3}{2} ) ^{2} - 2} d x

使用换元积分法

= \int \frac{- 2 u + 9}{2 4 u ^{2} - 2} d u

提出常数:

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

= \frac{1}{2} \cdot \int \frac{- 2 u + 9}{4 u ^{2} - 2} d u

乘开

\frac{- 2 u + 9}{4 u ^{2} - 2} : - \frac{2 u}{4 u ^{2} - 2} + \frac{9}{4 u ^{2} - 2}

= \frac{1}{2} \cdot \int - \frac{2 u}{4 u ^{2} - 2} + \frac{9}{4 u ^{2} - 2} d u

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

= \frac{1}{2} (- \int \frac{2 u}{4 u ^{2} - 2} d u + \int \frac{9}{4 u ^{2} - 2} d u)

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

\int \frac{9}{4 u ^{2} - 2} d u = \frac{9}{2} ln 2 u^{2} - 1 + 2 u

= \frac{1}{2} (- \frac{1}{2} 4 u^{2} - 2 + \frac{9}{2} ln 2 u^{2} - 1 + 2 u)

u = x - \frac{3}{2}

代回

= \frac{1}{2} - \frac{1}{2} 4 (x - \frac{3}{2})^{2} - 2 + \frac{9}{2} ln 2 (x - \frac{3}{2})^{2} - 1 + 2 (x - \frac{3}{2})

化简

\frac{1}{2} - \frac{1}{2} 4 (x - \frac{3}{2})^{2} - 2 + \frac{9}{2} ln 2 (x - \frac{3}{2})^{2} - 1 + 2 (x - \frac{3}{2}) : \frac{1}{4} (- 4 x^{2} - 12 x + 7 + 9 ln 2 (x^{2} - 3 x + \frac{9}{4}) - 1 + 2 (x - \frac{3}{2}))

= \frac{1}{4} (- 4 x^{2} - 12 x + 7 + 9 ln 2 (x^{2} - 3 x + \frac{9}{4}) - 1 + 2 (x - \frac{3}{2}))

解答补常数

= \frac{1}{4} (- 4 x^{2} - 12 x + 7 + 9 ln 2 (x^{2} - 3 x + \frac{9}{4}) - 1 + 2 (x - \frac{3}{2})) + C

作图

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