求不定积分 f[x^(3/2)]/(1+x) dx

答：f是积分号吧?
设t=√x,dx=d(t^2)=2tdt
∫ [ (x^(3/2) ] /(1+x) dx
=∫ (t^3) /(1+t^2) *2t dt
= 2 ∫ (t^4) /(1+t^2) dt
=2 ∫ (t^2+1-1)^2 /(1+t^2) dt
=2 ∫ (1+t^2) -2 +1/(1+t^2) dt
=2 ∫ (t^2-1) +1/(1+t^2) dt
=(2/3)*t^3 -2t +2arctant+C
=(2/3)*x^(3/2)-2√x+2arctan(√x)+C