令 x=(1/2)sit,則
I =∫[(x-1)/√(1-4x^2)]dx = (1/2)∫[(1/2)sint-1]dt = -(1/2)[(1/2)cost+t]+C
= -(1/4)[√(1-4x^2)+2arcsin(2x)]+C∫(1+x)/√(1-4x^2)dx 題目發(fā)錯了 你看看對不對 我做出來是-(1/4)[√(1-4x^2)-2arcsin(2x)]+C對的。
對于 ∫(1+x)/√(1-4x^2)dx
令 x=(1/2)sit,則
I =∫[(1+x)/√(1-4x^2)]dx = (1/2)∫[1+(1/2)sint]dt
= (1/2)[t-(1/2)cost]+C = (1/4)(2t-cost)+C

= (1/4)[2arcsin(2x)-√(1-4x^2)]+C.