1除以（x乘根号1-x^2）的不定积分

1除以（x乘根号1-x^2）的不定积分

这样的题一般用三角代换。令x=sint,dx=costdt
∫dx/[x√(1-x^2)]=∫costdt/[sintcost]=∫1/sint*dt=∫sint/(sint)^2*dt=∫sint/[1-(cost)^2]*dt
=-∫d(cost)/[1-(cost)^2]=-1/2*∫[1/(cost+1)-1/(cost-1)]d(cost)
=-1/2*(ln|cost+1|-ln|cost-1)+C
=-1/2*ln|(cost+1)/(cost-1)|+C
=-ln|(sint/(cost-1)|+C
=-ln|x/[√(1-x^2)-1]|+C
=ln|[√(1-x^2)-1]/x|+C

不知道

表达式不够明确，可能被理解为两种情形：x^2(√x)/(1-x) 或 (x^2)*√[x/(1-x)]； 如是第一种情形积分：设t=√x，则dx=2tdt； ∫[x^2(√x)/(1-x)]dx=∫[2t^6/(1-t^2]dt=-2∫t^4dt-2∫t^2dt-2∫dt+2∫[dt/(1-t^2) =-(2/5)t^5-(2/3)t^3-2t+(1/2)ln[(1+t)/(1-t)]+c =-(2/5)x^2√x-(2/3)x√x-2√x+(1/2)ln|(1+√x)/(1-√x)|+c； 如是第二种情形积分，有些麻烦：设t=√[x/(1-x)]，x=t^2/(1+t^2)，dx=2tdt/(1+t^2)^2； ∫x^2*√[x/(1-x)]dx=∫[t^2/(1+t^2)]^2*t*2tdt/(1+t^2)^2=∫[2t^6/(1+t^2)^4]dt； 再设tan u=t，则dt=du/(cosu)^2； 原积分=∫[2(tanu)^6/(1+(tanu)^2)^4] du/(cosu)^2=∫2(sinu)^6du=(1/4)∫(1-cos2u)^3 du =(1/4)∫[1-3cos2u+3(cos2u)^2-(cos2u)^3]du=u/4-(3/8)sin(2u)+[3u/2+(3/8)sin(4u)]-[(sin2u)/2-(sin2u)^3/6] =7u/4-(7/8)sin2u+(3/8)sin(4u)+(sin2u)^3/6+c 将u=arctan√[x/(1-x)]代入上式即得最后结果；

∫dx/[x√(1-x^2)] = ∫dx/[x^2 *√(1-1/x^2)] =-∫d(1/x)/√[1-(1/x)^2] =arccos(1/x)+C