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受欢迎的微积分 >

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

解答

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

解答

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

求解步骤

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

提出常数:

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

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

乘开

(x - 1)^{2} + 4 : x^{2} - 2 x + 5

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

对

x^{2} - 2 x + 5

配方:

(x - 1)^{2} + 4

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

使用换元积分法

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

乘开

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

= - 3 \cdot \int \frac{u}{u ^{2} + 4} + \frac{1}{u ^{2} + 4} 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

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

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

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

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

u = x - 1

代回

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

化简

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

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

解答补常数

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

作图

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