x根号下1一x平方dx-搜答案-搜搜考试网

xdx/(1-x*x)^(1/2)=-1/2*d(1-x*x)/(1-x*x)^(1/2)再问：我也是这样算的最后是负一但答案是1

设x=sint,dx=costdt,(以下省略积分符号）原式=[(sint)^2/cost]costdt=(sint)^2dt=(1-cos2t)/2*dt=1/2[dt-cos2tdt)=1/2t-

[x√(1-y²)]dx+[y√(1-x²)]dy=0[y√(1-x²)]dy＝-[x√(1-y²)]dx分离变量得ydy/√(1-y²)=-xdx/

∫1/[x√(1-ln²x)]dx=∫1/√(1-ln²x)d(lnx)=arcsin(lnx)+C公式：∫dx/√(a²-x²)=arcsin(x/a)+C

令x＝siny,则：√（1－x^2）＝√［1－（siny）^2］＝cosy,　y＝arcsinx,　dx＝cosydy.原式＝∫［cosy/（siny＋cosy）］dy　　＝∫｛cosy（cosy－s

x(1-y^2)^(1/2)dx+y(1-x^2)^(1/2)dy=0,|x|

原式=∫1/(1-x)(1+x)dx=1/2∫[1/(1-x)+1/(1+x)]dx=1/2[-ln|1-x|+ln|1+x|]+c=1/2ln|(1+x)/(1-x)|+c啊,原来有根号啊应该是ar

∫(-2→-1)√(3-4x-x^2)dx=∫(-2→-1)√[7-(x+2)^2]dxx+2=√7sinθ、dx=√7cosθdθθ∈[0,arcsin(1/√7)]=∫(√7cosθ)(√7cos

∫dx/[x^2√(1+x^2)]换元,x=tant=∫d(tant)/[tan^2t√(1+tan^2)]=∫(dt/cos^2t)/[tan^2t/cost]=∫dt/cost*tan^2t=∫c

由∫10²/√（1-x²）dx令t=cosx,dx=-sinxdt∴∫10²/√（1-cos²t）（-sintdt）=-∫100dt=-100t+C=-100a

x/Sqrt[1+x^2]+ln(x+Sqrt[1+x^2])

∫1/[x√(1-x²)]dx=∫1/[x*√[x²(1/x²-1)]dx=∫1/[x*|x|*√(1/x²-1)]dx=∫1/[x²√(1/x