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受欢迎的微积分 >

integral of 1/((u^2+9)^{5/2)}

解答

\int \frac{1}{( u ^{2} + 9 ) ^{\frac{5}{2}}} d u

解答

\frac{2 u ^{3} + 27 u}{243 ( u ^{2} + 9 ) ^{\frac{3}{2}}} + C

求解步骤

\int \frac{1}{( u ^{2} + 9 ) ^{\frac{5}{2}}} d u

使用三角换元法

= \int \frac{1}{81 sec ^{3} ( v )} d v

提出常数:

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

= \frac{1}{81} \cdot \int \frac{1}{sec ^{3} ( v )} d v

使用三角恒等式改写

= \frac{1}{81} \cdot \int (1 - sin^{2} (v)) cos (v) d v

使用换元积分法

= \frac{1}{81} \cdot \int 1 - w^{2} d w

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}{81} (\int 1 d w - \int w^{2} d w)

\int 1 d w = w

\int w^{2} d w = \frac{w ^{3}}{3}

= \frac{1}{81} (w - \frac{w ^{3}}{3})

代回

= \frac{1}{81} (sin (arctan (\frac{1}{3} u)) - \frac{sin ^{3} ( arctan ( \frac{1}{3} u ) )}{3})

化简

\frac{1}{81} (sin (arctan (\frac{1}{3} u)) - \frac{sin ^{3} ( arctan ( \frac{1}{3} u ) )}{3}) : \frac{2 u ^{3} + 27 u}{243 ( u ^{2} + 9 ) ^{\frac{3}{2}}}

= \frac{2 u ^{3} + 27 u}{243 ( u ^{2} + 9 ) ^{\frac{3}{2}}}

解答补常数

= \frac{2 u ^{3} + 27 u}{243 ( u ^{2} + 9 ) ^{\frac{3}{2}}} + C

作图

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