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受欢迎的微积分 >

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

解答

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

解答

- \frac{1}{4} (x - 2)^{3} - x^{2} + 4 x + 6 arcsin (\frac{1}{2} x - 1) - 3 sin (2 arcsin (\frac{1}{2} x - 1)) + C

求解步骤

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

使用换元积分法

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

使用三角换元法

= \int 16 sin^{4} (v) d v

提出常数:

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

= 16 \cdot \int sin^{4} (v) d v

使用降次积分法

= 16 (- \frac{cos ( v ) sin ^{3} ( v )}{4} + \frac{3}{4} \cdot \int sin^{2} (v) d v)

\int sin^{2} (v) d v = \frac{1}{2} (v - \frac{1}{2} sin (2 v))

= 16 (- \frac{cos ( v ) sin ^{3} ( v )}{4} + \frac{3}{4} \cdot \frac{1}{2} (v - \frac{1}{2} sin (2 v)))

代回

= 16 (- \frac{cos ( arcsin ( \frac{1}{2} ( x - 2 ) ) ) sin ^{3} ( arcsin ( \frac{1}{2} ( x - 2 ) ) )}{4} + \frac{3}{4} \cdot \frac{1}{2} (arcsin (\frac{1}{2} (x - 2)) - \frac{1}{2} sin (2 arcsin (\frac{1}{2} (x - 2)))))

化简

16 (- \frac{cos ( arcsin ( \frac{1}{2} ( x - 2 ) ) ) sin ^{3} ( arcsin ( \frac{1}{2} ( x - 2 ) ) )}{4} + \frac{3}{4} \cdot \frac{1}{2} (arcsin (\frac{1}{2} (x - 2)) - \frac{1}{2} sin (2 arcsin (\frac{1}{2} (x - 2))))) : - \frac{1}{4} (x - 2)^{3} - x^{2} + 4 x + 6 arcsin (\frac{1}{2} x - 1) - 3 sin (2 arcsin (\frac{1}{2} x - 1))

= - \frac{1}{4} (x - 2)^{3} - x^{2} + 4 x + 6 arcsin (\frac{1}{2} x - 1) - 3 sin (2 arcsin (\frac{1}{2} x - 1))

解答补常数

= - \frac{1}{4} (x - 2)^{3} - x^{2} + 4 x + 6 arcsin (\frac{1}{2} x - 1) - 3 sin (2 arcsin (\frac{1}{2} x - 1)) + C

作图

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