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受欢迎的微积分 >

integral of (-2x^2)/(sqrt(4x-x^2))

解答

\int \frac{- 2 x ^{2}}{4 x - x ^{2}} d x

解答

- 2 (6 arcsin (\frac{1}{2} x - 1) - sin (2 arcsin (\frac{1}{2} x - 1)) - 4 - x^{2} + 4 x) + C

求解步骤

\int \frac{- 2 x ^{2}}{4 x - x ^{2}} d x

提出常数:

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

= - 2 \cdot \int \frac{x ^{2}}{4 x - x ^{2}} d x

对

4 x - x^{2}

配方:

- (x - 2)^{2} + 4

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

使用换元积分法

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

使用三角换元法

= - 2 \cdot \int (2 sin (v) + 2)^{2} d v

乘开

(2 sin (v) + 2)^{2} : 4 sin^{2} (v) + 8 sin (v) + 4

= - 2 \cdot \int 4 sin^{2} (v) + 8 sin (v) + 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

= - 2 (\int 4 sin^{2} (v) d v + \int 8 sin (v) d v + \int 4 d v)

\int 4 sin^{2} (v) d v = 2 v - sin (2 v)

\int 8 sin (v) d v = - 8 cos (v)

\int 4 d v = 4 v

= - 2 (2 v - sin (2 v) - 8 cos (v) + 4 v)

代回

= - 2 (2 arcsin (\frac{1}{2} (x - 2)) - sin (2 arcsin (\frac{1}{2} (x - 2))) - 8 cos (arcsin (\frac{1}{2} (x - 2))) + 4 arcsin (\frac{1}{2} (x - 2)))

化简

- 2 (2 arcsin (\frac{1}{2} (x - 2)) - sin (2 arcsin (\frac{1}{2} (x - 2))) - 8 cos (arcsin (\frac{1}{2} (x - 2))) + 4 arcsin (\frac{1}{2} (x - 2))) : - 2 (6 arcsin (\frac{1}{2} x - 1) - sin (2 arcsin (\frac{1}{2} x - 1)) - 4 - x^{2} + 4 x)

= - 2 (6 arcsin (\frac{1}{2} x - 1) - sin (2 arcsin (\frac{1}{2} x - 1)) - 4 - x^{2} + 4 x)

解答补常数

= - 2 (6 arcsin (\frac{1}{2} x - 1) - sin (2 arcsin (\frac{1}{2} x - 1)) - 4 - x^{2} + 4 x) + C

作图

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