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受欢迎的微积分 >

integral of (2x+3)/(sqrt(4x^2-4x+5))

解答

\int \frac{2 x + 3}{4 x ^{2} - 4 x + 5} d x

解答

\frac{1}{2} (4 x^{2} - 4 x + 5 + 4 ln (\frac{2 x + 4 x ^{2} - 4 x + 5 - 1}{2})) + C

求解步骤

\int \frac{2 x + 3}{4 x ^{2} - 4 x + 5} d x

使用换元积分法

= \int \frac{u}{2 u ^{2} - 8 u + 20} 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{u}{u ^{2} - 8 u + 20} d u

对

u^{2} - 8 u + 20

配方:

(u - 4)^{2} + 4

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

使用换元积分法

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

乘开

\frac{v + 4}{v ^{2} + 4} : \frac{v}{v ^{2} + 4} + \frac{4}{v ^{2} + 4}

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

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{v}{v ^{2} + 4} d v + \int \frac{4}{v ^{2} + 4} d v)

\int \frac{v}{v ^{2} + 4} d v = v^{2} + 4

\int \frac{4}{v ^{2} + 4} d v = 4 ln (\frac{1}{2} v + 4 + v^{2})

= \frac{1}{2} (v^{2} + 4 + 4 ln (\frac{1}{2} v + 4 + v^{2}))

代回

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

化简

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

= \frac{1}{2} (4 x^{2} - 4 x + 5 + 4 ln (\frac{2 x + 4 x ^{2} - 4 x + 5 - 1}{2}))

解答补常数

= \frac{1}{2} (4 x^{2} - 4 x + 5 + 4 ln (\frac{2 x + 4 x ^{2} - 4 x + 5 - 1}{2})) + C

作图

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