一道不定积分,求助
1个回答
2017-11-06
展开全部
[x^3/√(x^2+1)]dx =∫[x^2/√(x^2+1)]xdx =1/2∫[x^2/√(x^2+1)]d(x^2+1) =1/2∫*1/2*x^2d√(x^2+1) =1/4∫x^2d√(x^2+1) =1/4(x^2*√(x^2+1)-∫√(x^2+1)dx^2) =1/4(x^2*√(x^2+1)-∫√(x^2+1)d(x^2+1)) =1/4(x^2*√(x^2+1)-3/2√(x^2+1)^3)+C =1/4x^2*√(x^2+
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询