Question

这个定积分怎么做

这个定积分怎么做这个定积分怎么做？谢谢

Answer 1

我这个结果是完全正确的。

Answer 2

∫(0->3) x^2/(x^2-3x+3)^2 dx
=∫(0->3) dx/(x^2-3x+3) +∫(0->3) (3x-3)/(x^2-3x+3)^2 dx
=∫(0->3) dx/(x^2-3x+3) +(3/2)∫(0->3) (2x-3)/(x^2-3x+3)^2 dx
-(3/2)∫(0->3) dx/(x^2-3x+3)^2
= -(3/2)[1/(x^2-3x+3)]|(0->3)+∫(0->3) dx/(x^2-3x+3) -(3/2)∫(0->3) dx/(x^2-3x+3)^2
=0+∫(0->3) dx/(x^2-3x+3) -(3/2)∫(0->3) dx/(x^2-3x+3)^2
= 0+(4√3/9)π -(3/2)[ (4√3/9) ( 2π/3 - 1/2 ) ]
=3
consider
x^2-3x+3 = (x- 3/2)^2 + 3/4
let
x-3/2 = (√3/2) tanu
dx= (√3/2) (secu)^2 du
x=0, u=-π/3
x=3, u=π/3
∫(0->3) dx/(x^2-3x+3)
=∫(-π/3->π/3) (√3/2) (secu)^2 du/[ (3/4)(secu)^2]
=(2√3/3) ∫(-π/3->π/3) du
=(4√3/9)π
∫(0->3) dx/(x^2-3x+3)^2
=∫(-π/3->π/3) (√3/2) (secu)^2 du/[ (3/4)(secu)^2]^2
=(8√3/9) ∫(-π/3->π/3) (cosu)^2 du
=(4√3/9) ∫(-π/3->π/3) (1+cos2u) du
=(4√3/9) [u+(1/2)cos2u] |(-π/3->π/3)
=(4√3/9) ( 2π/3 - 1/2 )