Question

换元积分法 凑微分

Answer 1

∫[(1-x)/√(9-4x^2) ] dx
=(1/8) ∫ d(9-4x^2)/√(9-4x^2) + ∫ dx/√(9-4x^2)
=(1/4) √(9-4x^2) + ∫ dx/√(9-4x^2)
let
2x = 3sina
2dx = 3cosa da
∫ dx/√(9-4x^2)
=∫ [1/ (3cosa)] .(3cosa/2) da
=(1/2)a + C
=(1/2)arcsin(2x/3) +C

∫[(1-x)/√(9-4x^2) ] dx
=(1/4) √(9-4x^2) + ∫ dx/√(9-4x^2)
=(1/4) √(9-4x^2) + (1/2)arcsin(2x/3) +C

Answer 2

原式=∫dx/√(9-4x^2)+1/8∫(-8xdx)/√(9-4x^2)
=1/3∫dx/√[1-(2/3x)^2]+1/8∫d(9-4x^2)/√(9-4x^2)
=1/2∫d(2/3x)/√[1-(2/3x)^2]+1/8∫(9-4x^2)^(-1/2)d(9-4x^2)
=1/2*arcsin(2/3 x)+1/4√(9-4x^2)+C

Answer 3

分成两项来做；
原式=∫1/√(9-4x²)dx - ∫x/√(9-4x²)dx
凑微分：第一项：t=(3/2)x, 第二项里面t=9-4x²