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受欢迎的微积分 >

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

解答

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

解答

- \frac{( x ^{2} + 2 ) - x ^{2} + 4}{24 x ^{3}} + C

求解步骤

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

使用三角换元法

= \int \frac{1}{16 sin ^{4} ( u )} d u

提出常数:

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

= \frac{1}{16} \cdot \int \frac{1}{sin ^{4} ( u )} d u

使用三角恒等式改写

= \frac{1}{16} \cdot \int csc^{4} (u) d u

使用降次积分法

= \frac{1}{16} (- \frac{cos ( u ) csc ^{3} ( u )}{3} + \frac{2}{3} \cdot \int csc^{2} (u) d u)

\int csc^{2} (u) d u = - cot (u)

= \frac{1}{16} (- \frac{cos ( u ) csc ^{3} ( u )}{3} + \frac{2}{3} (- cot (u)))

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

代回

= \frac{1}{16} (- \frac{cos ( arcsin ( \frac{1}{2} x ) ) csc ^{3} ( arcsin ( \frac{1}{2} x ) )}{3} + \frac{2}{3} (- cot (arcsin (\frac{1}{2} x))))

化简

\frac{1}{16} (- \frac{cos ( arcsin ( \frac{1}{2} x ) ) csc ^{3} ( arcsin ( \frac{1}{2} x ) )}{3} + \frac{2}{3} (- cot (arcsin (\frac{1}{2} x)))) : - \frac{( x ^{2} + 2 ) - x ^{2} + 4}{24 x ^{3}}

= - \frac{( x ^{2} + 2 ) - x ^{2} + 4}{24 x ^{3}}

解答补常数

= - \frac{( x ^{2} + 2 ) - x ^{2} + 4}{24 x ^{3}} + C

作图

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\int \frac{e ^{5 x}}{5} d x

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