Question

求不定积分 f[x^(3/2)]/(1+x) dx？

Answer 1

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=(...,1,f[x^(3/2)]/(1+x) dx
=fxx^(1/2)/(1+x)dx
=f(1+x-1)/(1+x) *x^（1/2)dx
=f(1-1/(1+x))x^(1/2)DX
=f(x^(1/2)dx-x^(1/2)/(1+x)dx
由于x>0 可以x=tan^2t
dx=2tantsec^2xtdt
后者=ftant/sec^2t*2...,0,求不定积分 f[x^(3/2)]/(1+x) dx
f[x^(3/2)]/(1+x) dx