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受欢迎的微积分 >

integral of (sin(arctan(9x)))/(1+81x^2)

解答

\int \frac{sin ( arctan ( 9 x ) )}{1 + 81 x ^{2}} d x

解答

- \frac{1}{9 1 + 81 x ^{2}} + C

求解步骤

\int \frac{sin ( arctan ( 9 x ) )}{1 + 81 x ^{2}} d x

使用换元积分法

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

使用换元积分法

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

提出常数:

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

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

化简

v v : v^{\frac{3}{2}}

= \frac{1}{9} \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{\frac{3}{2}}} d v

使用幂法则

= \frac{1}{9} \cdot \frac{1}{2} (- \frac{2}{v})

代回

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

化简

\frac{1}{9} \cdot \frac{1}{2} - \frac{2}{1 + ( 9 x ) ^{2}} : - \frac{1}{9 1 + 81 x ^{2}}

= - \frac{1}{9 1 + 81 x ^{2}}

解答补常数

= - \frac{1}{9 1 + 81 x ^{2}} + C

作图

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