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受欢迎的微积分 >

integral of 1/((9+x^2)^3)

解答

\int \frac{1}{( 9 + x ^{2} ) ^{3}} d x

解答

\frac{x ^{4} arctan ( \frac{x}{3} ) + 3 x ^{3} + 18 x ^{2} arctan ( \frac{x}{3} ) + 45 x + 81 arctan ( \frac{x}{3} )}{648 ( x ^{2} + 9 ) ^{2}} + C

求解步骤

\int \frac{1}{( 9 + x ^{2} ) ^{3}} d x

使用指数法则:

\frac{1}{a ^{b}} = a^{- b}

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

= \int (9 + x^{2})^{- 3} d x

使用换元积分法

= \int \frac{1}{243 ( u ^{2} + 1 ) ^{3}} d u

提出常数:

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

= \frac{1}{243} \cdot \int \frac{1}{( u ^{2} + 1 ) ^{3}} d u

使用降次积分法

= \frac{1}{243} (\frac{3}{4} \cdot \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u + \frac{u}{2 \cdot 2 ( u ^{2} + 1 ) ^{2}})

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

= \frac{1}{243} (\frac{3}{4} \cdot \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}) + \frac{u}{2 \cdot 2 ( u ^{2} + 1 ) ^{2}})

u = \frac{x}{3}

代回

= \frac{1}{243} \frac{3}{4} \cdot \frac{1}{2} (arctan (\frac{x}{3}) + \frac{\frac{x}{3}}{1 + ( \frac{x}{3} ) ^{2}}) + \frac{\frac{x}{3}}{2 \cdot 2 ( ( \frac{x}{3} ) ^{2} + 1 ) ^{2}}

化简

\frac{1}{243} \frac{3}{4} \cdot \frac{1}{2} (arctan (\frac{x}{3}) + \frac{\frac{x}{3}}{1 + ( \frac{x}{3} ) ^{2}}) + \frac{\frac{x}{3}}{2 \cdot 2 ( ( \frac{x}{3} ) ^{2} + 1 ) ^{2}} : \frac{x ^{4} arctan ( \frac{x}{3} ) + 3 x ^{3} + 18 x ^{2} arctan ( \frac{x}{3} ) + 45 x + 81 arctan ( \frac{x}{3} )}{648 ( x ^{2} + 9 ) ^{2}}

= \frac{x ^{4} arctan ( \frac{x}{3} ) + 3 x ^{3} + 18 x ^{2} arctan ( \frac{x}{3} ) + 45 x + 81 arctan ( \frac{x}{3} )}{648 ( x ^{2} + 9 ) ^{2}}

解答补常数

= \frac{x ^{4} arctan ( \frac{x}{3} ) + 3 x ^{3} + 18 x ^{2} arctan ( \frac{x}{3} ) + 45 x + 81 arctan ( \frac{x}{3} )}{648 ( x ^{2} + 9 ) ^{2}} + C

作图

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