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受欢迎的微积分 >

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

解答

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

解答

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

求解步骤

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

对

x^{2} - 4 x + 3

配方:

(x - 2)^{2} - 1

= \int (x - 2)^{2} - 1 d x

使用换元积分法

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

使用三角换元法

= \int tan^{2} (v) sec (v) d v

使用三角恒等式改写

= \int (- 1 + sec^{2} (v)) sec (v) d v

乘开

(- 1 + sec^{2} (v)) sec (v) : - sec (v) + sec^{3} (v)

= \int - sec (v) + sec^{3} (v) 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

= - \int sec (v) d v + \int sec^{3} (v) d v

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

\int sec^{3} (v) d v = \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

= - ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

代回

= - ln ∣ tan (arcsec (x - 2)) + sec (arcsec (x - 2)) ∣ + \frac{1}{2} sec (arcsec (x - 2)) tan (arcsec (x - 2)) + \frac{1}{2} ln ∣ tan (arcsec (x - 2)) + sec (arcsec (x - 2)) ∣

化简

- ln ∣ tan (arcsec (x - 2)) + sec (arcsec (x - 2)) ∣ + \frac{1}{2} sec (arcsec (x - 2)) tan (arcsec (x - 2)) + \frac{1}{2} ln ∣ tan (arcsec (x - 2)) + sec (arcsec (x - 2)) ∣ : - ln x^{2} - 4 x + 3 + x - 2 + \frac{1}{2} (x - 2) x^{2} - 4 x + 3 + \frac{1}{2} ln x^{2} - 4 x + 3 + x - 2

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

解答补常数

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

作图

流行的例子

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\int \frac{y}{3 y + 1} d y

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