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受欢迎的微积分 >

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

解答

\int \frac{1}{x ^{4} x ^{2} + 3} d x

解答

\frac{1}{9} - \frac{( 3 + x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}} + \frac{3 + x ^{2}}{x} + C

求解步骤

\int \frac{1}{x ^{4} x ^{2} + 3} d x

使用三角换元法

= \int \frac{sec ( u )}{9 tan ^{4} ( u )} d u

提出常数:

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

= \frac{1}{9} \cdot \int \frac{sec ( u )}{tan ^{4} ( u )} d u

使用三角恒等式改写

= \frac{1}{9} \cdot \int \frac{( 1 - sin ^{2} ( u ) ) cos ( u )}{sin ^{4} ( u )} d u

使用换元积分法

= \frac{1}{9} \cdot \int \frac{1 - v ^{2}}{v ^{4}} d v

乘开

\frac{1 - v ^{2}}{v ^{4}} : \frac{1}{v ^{4}} - \frac{1}{v ^{2}}

= \frac{1}{9} \cdot \int \frac{1}{v ^{4}} - \frac{1}{v ^{2}} 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

= \frac{1}{9} (\int \frac{1}{v ^{4}} d v - \int \frac{1}{v ^{2}} d v)

\int \frac{1}{v ^{4}} d v = - \frac{1}{3 v ^{3}}

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

= \frac{1}{9} (- \frac{1}{3 v ^{3}} - (- \frac{1}{v}))

代回

= \frac{1}{9} - \frac{1}{3 sin ^{3} ( arctan ( \frac{1}{3} x ) )} - - \frac{1}{sin ( arctan ( \frac{1}{3} x ) )}

化简

\frac{1}{9} - \frac{1}{3 sin ^{3} ( arctan ( \frac{1}{3} x ) )} - - \frac{1}{sin ( arctan ( \frac{1}{3} x ) )} : \frac{1}{9} - \frac{( 3 + x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}} + \frac{3 + x ^{2}}{x}

= \frac{1}{9} - \frac{( 3 + x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}} + \frac{3 + x ^{2}}{x}

解答补常数

= \frac{1}{9} - \frac{( 3 + x ^{2} ) ^{\frac{3}{2}}}{3 x ^{3}} + \frac{3 + x ^{2}}{x} + C

作图

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