Question

求积分，最好给过程

Answer 1

let
x= sinu
dx = cosu du
x=0, u=0
x=√3/2 , u=π/3
∫(-√3/2->√3/2) dx/(1-2x^2+x^4)

=2∫(0->√3/2) dx/(1-2x^2+x^4)
=2∫(0->√3/2) dx/(1-x^2)^2
=2∫(0->π/3) ( cosu du )/(cosu )^4
=2∫(0->π/3) (secu)^3 du
=2[√3 + (1/2)ln(2 +√3)]
=2√3 + ln(2 +√3)
------------------------
∫(0->π/3) (secu)^3 du
=∫(0->π/3) secu.dtanu
=[secu.tanu]|(0->π/3) - ∫(0->π/3) (tanu)^2.secu du
=2√3 - ∫(0->π/3) [ (secu)^2-1].secu du
2∫(0->π/3) (secu)^3 du =2√3 + ∫(0->π/3) secu du
∫(0->π/3) (secu)^3 du
=√3 + (1/2)∫(0->π/3) secu du
=√3 + (1/2)[ln|secu +tanu|] |(0->π/3)
=√3 + (1/2)ln(2 +√3)

Answer 2

字写得不是很好

Answer 3

定积分偶倍奇零
=2∫(0.√3/2)1/(1-x²)²dx
令x=sinu
=2∫(0.π/3)1/(cos²u)²dsinu
=2∫sec³udu
=2∫secudtanu
=secutanu+ln|secu+tanu|
=2√3+ln(2+√3)