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受欢迎的微积分 >

integral of 1/([(x^2+9)^2])

解答

\int \frac{1}{[ ( x ^{2} + 9 ) ^{2} ]} d x

解答

\frac{1}{54} (arctan (\frac{x}{3}) + \frac{1}{2} sin (2 arctan (\frac{x}{3}))) + C

求解步骤

\int \frac{1}{[ ( x ^{2} + 9 ) ^{2} ]} d x

化简

= \int \frac{1}{( x ^{2} + 9 ) ^{2}} d x

使用换元积分法

= \int \frac{1}{27 ( u ^{2} + 1 ) ^{2}} d u

提出常数:

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

= \frac{1}{27} \cdot \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u

使用三角换元法

= \frac{1}{27} \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

使用三角恒等式改写

= \frac{1}{27} \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

= \frac{1}{27} \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

= \frac{1}{27} \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)

= \frac{1}{27} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

代回

= \frac{1}{27} \cdot \frac{1}{2} (arctan (\frac{x}{3}) + \frac{1}{2} sin (2 arctan (\frac{x}{3})))

化简

\frac{1}{27} \cdot \frac{1}{2} (arctan (\frac{x}{3}) + \frac{1}{2} sin (2 arctan (\frac{x}{3}))) : \frac{1}{54} (arctan (\frac{x}{3}) + \frac{1}{2} sin (2 arctan (\frac{x}{3})))

= \frac{1}{54} (arctan (\frac{x}{3}) + \frac{1}{2} sin (2 arctan (\frac{x}{3})))

解答补常数

= \frac{1}{54} (arctan (\frac{x}{3}) + \frac{1}{2} sin (2 arctan (\frac{x}{3}))) + C

作图

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