求不定积分∫x/√(4-x²) dx

问题描述:

求不定积分∫x/√(4-x²) dx

∫x/√(4-x^2)dx
=-(1/2)∫d(4-x^2)/√(4-x^2)
=-(1/4)√(4-x^2)+C=-(1/2)∫d(4-x^2)/√(4-x^2)这一步怎么求的呢,求解,谢谢xdx=d(x^2)=-(1/2)d(-x^2)=-(1/2)d(4-x^2)你得从求导数来看:[-(1/2)(4-x^2)]'=x,所以d[-(1/2)(4-x^2)]'=dx