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受欢迎的微积分 >

integral of (6x-5)/(sqrt(3x^2+4x+1))

解答

\int \frac{6 x - 5}{3 x ^{2} + 4 x + 1} d x

解答

\frac{1}{3} (36 x^{2} + 48 x + 12 - 9 ln (\frac{6 x + 36 x ^{2} + 48 x + 12 + 4}{2})) + C

求解步骤

\int \frac{6 x - 5}{3 x ^{2} + 4 x + 1} d x

使用换元积分法

= \int \frac{u}{3 u ^{2} + 18 u + 77} d u

提出常数:

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

= \frac{1}{3} \cdot \int \frac{u}{u ^{2} + 18 u + 77} d u

对

u^{2} + 18 u + 77

配方:

(u + 9)^{2} - 4

= \frac{1}{3} \cdot \int \frac{u}{( u + 9 ) ^{2} - 4} d u

使用换元积分法

= \frac{1}{3} \cdot \int \frac{v - 9}{v ^{2} - 4} d v

乘开

\frac{v - 9}{v ^{2} - 4} : \frac{v}{v ^{2} - 4} - \frac{9}{v ^{2} - 4}

= \frac{1}{3} \cdot \int \frac{v}{v ^{2} - 4} - \frac{9}{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}{3} (\int \frac{v}{v ^{2} - 4} d v - \int \frac{9}{v ^{2} - 4} d v)

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

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

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

代回

= \frac{1}{3} ((6 x - 5 + 9)^{2} - 4 - 9 ln (\frac{1}{2} (6 x - 5 + 9)^{2} - 4 + 6 x - 5 + 9))

化简

\frac{1}{3} ((6 x - 5 + 9)^{2} - 4 - 9 ln (\frac{1}{2} (6 x - 5 + 9)^{2} - 4 + 6 x - 5 + 9)) : \frac{1}{3} (36 x^{2} + 48 x + 12 - 9 ln (\frac{6 x + 36 x ^{2} + 48 x + 12 + 4}{2}))

= \frac{1}{3} (36 x^{2} + 48 x + 12 - 9 ln (\frac{6 x + 36 x ^{2} + 48 x + 12 + 4}{2}))

解答补常数

= \frac{1}{3} (36 x^{2} + 48 x + 12 - 9 ln (\frac{6 x + 36 x ^{2} + 48 x + 12 + 4}{2})) + C

作图

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