求不定积分∫[x√(4-x²)]dx_数学解答

求不定积分∫[x√(4-x²)]dx

人气：365 ℃　时间：2020-06-12 05:51:26

解答

原式=1/2∫根号下(4-x^2)dx^2=1/2∫根号下(4-t)dt=-1/2*2/3*(4-t)^(3/2)+C=-1/3*(4-x^2)^(3/2)+C