这题怎么做?…求完整解答,跪谢!!!
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∫(1-x)/√(9-4x^2) dx
=(1/8)∫(-8x)/√(9-4x^2) dx +∫dx/√(9-4x^2)
=(1/4)√(9-4x^2) + ∫dx/√(9-4x^2)
let
2x= 3siny
2dx =3cosy dy
∫dx/√(9-4x^2)
= (1/2)∫ dy
=(1/2)y + C
=(1/2)arsin(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)arsin(2x/3) + C
=(1/8)∫(-8x)/√(9-4x^2) dx +∫dx/√(9-4x^2)
=(1/4)√(9-4x^2) + ∫dx/√(9-4x^2)
let
2x= 3siny
2dx =3cosy dy
∫dx/√(9-4x^2)
= (1/2)∫ dy
=(1/2)y + C
=(1/2)arsin(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)arsin(2x/3) + C
追问
太感谢了!就是全是文字形式看起来有点小复杂
👍👍👍👍
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