请问下这道高数题怎么做？不定积分
展开全部
∫(1-x)/√(9-x^2)dx
=∫1/√(9-x^2)dx-∫x/√(9-x^2)dx
=∫1/√[1-(x/3)^2]d(x/3)+1/2∫1/√(9-x^2)d(9-x^2)
=arcsin(x/3)+1/2×1/(-1/2+1)(9-x^2)^(-1/2+1)+C
=arcsin(x/3)+(1/4)√(9-x^2)+C
=∫1/√(9-x^2)dx-∫x/√(9-x^2)dx
=∫1/√[1-(x/3)^2]d(x/3)+1/2∫1/√(9-x^2)d(9-x^2)
=arcsin(x/3)+1/2×1/(-1/2+1)(9-x^2)^(-1/2+1)+C
=arcsin(x/3)+(1/4)√(9-x^2)+C
本回答由提问者推荐
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
