Question

（1-x）^2/根号x 的不定积分,

Answer 1

设分母√x=t

则

x=t^2

dx=2tdt

则原式化为

∫(1-t^2)^2*2tdt/t

=2∫(1-t^2)^2dt

=2∫(1-2t^2+t^4)dt

=2t-4t^3/3+2t^5/5+C

=2√x-4√x^3 / 3 +2√x^5 / 5 +C

=2√x-4x√x /3+2x^2√x / 5 +C