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受欢迎的微积分 >

integral of (x^2-y^2)/((x^2+y^2)^2)

解答

\int \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} d y

解答

\frac{y}{y ^{2} + x ^{2}} + C

求解步骤

\int \frac{x ^{2} - y ^{2}}{( x ^{2} + y ^{2} ) ^{2}} d y

使用换元积分法

= \int \frac{x ( x ^{2} - x ^{2} u ^{2} )}{( x ^{2} + x ^{2} u ^{2} ) ^{2}} d u

提出常数:

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

= x \cdot \int \frac{x ^{2} - x ^{2} u ^{2}}{( x ^{2} + x ^{2} u ^{2} ) ^{2}} d u

分解

\frac{x ^{2} - x ^{2} u ^{2}}{( x ^{2} + x ^{2} u ^{2} ) ^{2}} : \frac{- ( u + 1 ) ( u - 1 )}{x ^{2} ( u ^{2} + 1 ) ^{2}}

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

提出常数:

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

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

化简

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

将

\frac{( u + 1 ) ( u - 1 )}{( u ^{2} + 1 ) ^{2}}

用部份分式展开:

\frac{1}{u ^{2} + 1} - \frac{2}{( u ^{2} + 1 ) ^{2}}

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

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

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

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

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

= x (- \frac{1}{x ^{2}}) (arctan (u) - (arctan (u) + \frac{u}{1 + u ^{2}}))

u = \frac{y}{x}

代回

= x (- \frac{1}{x ^{2}}) (arctan (\frac{y}{x}) - (arctan (\frac{y}{x}) + \frac{\frac{y}{x}}{1 + ( \frac{y}{x} ) ^{2}}))

化简

x (- \frac{1}{x ^{2}}) (arctan (\frac{y}{x}) - (arctan (\frac{y}{x}) + \frac{\frac{y}{x}}{1 + ( \frac{y}{x} ) ^{2}})) : \frac{y}{y ^{2} + x ^{2}}

= \frac{y}{y ^{2} + x ^{2}}

解答补常数

= \frac{y}{y ^{2} + x ^{2}} + C

作图

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