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受欢迎的微积分 >

integral of (x-1)/((x^2+4x+5)^2)

解答

\int \frac{x - 1}{( x ^{2} + 4 x + 5 ) ^{2}} d x

解答

- \frac{1}{2 ( x ^{2} + 4 x + 5 )} - \frac{3 ( arctan ( x + 2 ) ( x ^{2} + 4 x + 5 ) + x + 2 )}{2 ( x ^{2} + 4 x + 5 )} + C

求解步骤

\int \frac{x - 1}{( x ^{2} + 4 x + 5 ) ^{2}} d x

对

x^{2} + 4 x + 5

配方:

(x + 2)^{2} + 1

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

使用换元积分法

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

乘开

\frac{u - 3}{( u ^{2} + 1 ) ^{2}} : \frac{u}{( u ^{2} + 1 ) ^{2}} - \frac{3}{( u ^{2} + 1 ) ^{2}}

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

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

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

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

= - \frac{1}{2 ( u ^{2} + 1 )} - \frac{3}{2} (arctan (u) + \frac{u}{1 + u ^{2}})

u = x + 2

代回

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

化简

- \frac{1}{2 ( ( x + 2 ) ^{2} + 1 )} - \frac{3}{2} (arctan (x + 2) + \frac{x + 2}{1 + ( x + 2 ) ^{2}}) : - \frac{1}{2 ( x ^{2} + 4 x + 5 )} - \frac{3 ( arctan ( x + 2 ) ( x ^{2} + 4 x + 5 ) + x + 2 )}{2 ( x ^{2} + 4 x + 5 )}

= - \frac{1}{2 ( x ^{2} + 4 x + 5 )} - \frac{3 ( arctan ( x + 2 ) ( x ^{2} + 4 x + 5 ) + x + 2 )}{2 ( x ^{2} + 4 x + 5 )}

解答补常数

= - \frac{1}{2 ( x ^{2} + 4 x + 5 )} - \frac{3 ( arctan ( x + 2 ) ( x ^{2} + 4 x + 5 ) + x + 2 )}{2 ( x ^{2} + 4 x + 5 )} + C

作图

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