不定积分x^2乘根号下x/1-x

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；