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受欢迎的微积分 >

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

解答

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

解答

\frac{x}{4 4 - x ^{2}} + C

求解步骤

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

(4 - x^{2})^{3} = (4 - x^{2})^{\frac{3}{2}},

假定

(4 - x^{2}) \geq 0

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

使用三角换元法

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

提出常数:

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

= \frac{1}{4} \cdot \int \frac{1}{cos ^{2} ( u )} d u

使用积分常见定则:

\int \frac{1}{cos ^{2} ( u )} d u = tan (u)

= \frac{1}{4} tan (u)

u = arcsin (\frac{1}{2} x)

代回

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

化简

\frac{1}{4} tan (arcsin (\frac{1}{2} x)) : \frac{x}{4 4 - x ^{2}}

= \frac{x}{4 4 - x ^{2}}

解答补常数

= \frac{x}{4 4 - x ^{2}} + C

作图

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x \to \infty lim (1.0 1^{x})

(x^2-x+1)y^{''}-(x^2+x)y^'+(x+1)y=0

(x^{2} - x + 1) y^{^{''}} - (x^{2} + x) y^{^{'}} + (x + 1) y = 0

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