Question

求不定积分∫√(1-x)/√(1+x)xdx

求不定积分∫√(1-x)/√(1+x)xdx
可能我刚刚没有写清楚，题目是这样的：
∫［√（1-x²）/（1+x²）］xdx

Answer 1

简单计算一下即可，答案如图所示

Answer 2

∫√[(1-x)/(1+x)] dx/x^2 x=cosu dx=sinu √[(1-x)/(1+x)]=(1-cosu)/sinu 原式=∫(1-cosu)du/(cosu)^2 =∫du/(cosu)^2-∫cosudu/(1-sinu)(1+sinu) =tanu-∫dsinu/[(1-sinu)(1+sinu)] =tanu-ln[|1+sinu|/|cosu|] +C =√[(1/x^2)-1 ] -ln[|1+√(1-x^2)|/|x| ]+C ∫dx/(1+x^3) =∫dx/[(1+x)(1-x+x^2)]=(1/3)∫(1+x)dx/[x(1-x+x^2)]-(1/3)∫dx/[x(1+x)] =(1/3)∫xdx/[x^2(1-x+x^2)]+(1/3)∫dx/[(x-1/2)^2+3/4]-(1/3)ln[|1+x|/|x|] =(1/6)∫dx^2/(x^2(1-x+x^2)+... ...=(2/3√3)arctan(2x/√3-1/√3)-(1/3)ln|1+x|/|x| =(1/6)∫dx^2/[(x-1)(1-x+x^2]+(1/6)∫dx^2/[(x-1)x^2]+... =(1/3)∫xdx/[(x-1)(x^2-x+1)]+(1/3)∫dx/[x(x-1)]+... =(1/3)∫(x-1)dx/(x^2-x+1)-(1/3)∫dx/(x-1)+(1/3)∫dx/(x-1)-(1/3)∫dx/x+... =(1/6)ln|x^2-x+1|-(1/6)∫dx/[(x-1/2)^2+3/4]-(1/3)ln|x|+... =(1/6)ln|x^2-x+1|+(1/3√3)arctan(2x/√3-1/√3)-(1/3)ln|1+x|+C