Question

一道关于不定积分的高数题？

怎么用分部积分法求?感谢!

Answer 1

let
√x = sinu
dx = 2sinu.cosu du
∫ arcsin√x /√(1-x) dx
=∫ [u/cosu] [2sinu.cosu du]
=2∫ usinu du
=-2∫ u dcosu
=-2ucosu +2∫ cosu du
=-2ucosu +2sinu +C
=-2(arcsin√x) .√(1-x) +2√x +C