求不定积分x/[(x^3)(1-x)]^1/4不定积分的符号我就不写了
热心网友
I=∫xdx/[(x^3)(1-x)]^1/4=∫[x/(1-x)]^(1/4)dx,[x/（1-x）]^1/4=u==dx=4u^3du/(u^4+1)^2 ==I=4∫u^4du/(u^4+1)^2=∫ud(u^4)/(u^4+1)^2=-∫ud[1/(u^4+1)]=-u/(u^4+1)+∫du/(u^4+1)==-u/(u^4+1)+(1/2)∫(1+u^2)du/(u^4+1)+(1/2)∫(1-u^2)du/(u^4+1)==-u/(u^4+1)+(1/2)∫(1+u^(-2))du/(u^2+u^(-2))--(1/2)∫(1-u^(-2))du/(u^2+u^(-2))==-u/(u^4+1)+(1/2)∫d(u-u^(-1))/[(u-u^(-1))^2+2]--(1/2)∫d(u+u^(-1))/[(u+u^(-1))^2-2]==-u/(u^4+1)+[(1/2*2*√2)]ln[(u+u^(-1)-√2)/(u+u^(-1)+√2)]-(1/2*√2)arctg[(u-u^(-1))/√2],再代入[x/（1-x）]^1/4=u。
热心网友
求不定积分x/[(x^3)(1-x)]^1/4不定积分的符号我就不写了I=Integrate[x/(1-x)]^(1/4)再令[x/(1-x)]^(1/4)=u,则x=u^4/(u^4+1),dx=4u^3/(u^4+1)^2 du再代入计算，try it youself, it's very complicated!