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受欢迎的微积分 >

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

解答

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

解答

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

求解步骤

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

使用换元积分法

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

提出常数:

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

= 2 \cdot \int 4 - u^{2} d u

使用三角换元法

= 2 \cdot \int 4 cos^{2} (v) d v

提出常数:

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

= 2 \cdot 4 \cdot \int cos^{2} (v) d v

使用三角恒等式改写

= 2 \cdot 4 \cdot \int \frac{1 + cos ( 2 v )}{2} d v

提出常数:

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

= 2 \cdot 4 \cdot \frac{1}{2} \cdot \int 1 + cos (2 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

= 2 \cdot 4 \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

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

= 2 \cdot 4 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

代回

= 2 \cdot 4 \cdot \frac{1}{2} (arcsin (\frac{1}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} x)))

化简

2 \cdot 4 \cdot \frac{1}{2} (arcsin (\frac{1}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} x))) : 4 (arcsin (\frac{1}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} x)))

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

解答补常数

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

作图

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