请问32题怎么做
2个回答
展开全部
∫(1-x)/√(9-4x^2) dx
=(1/8)∫(-8x)/√(9-4x^2) dx +∫dx/√(9-4x^2)
=(1/4) √(9-4x^2) +∫dx/√(9-4x^2)
=(1/4) √(9-4x^2) +(1/2)arcsin(2x/3) + C
//
let
x=(3/2)sinu
dx=(3/2)cosu du
∫dx/√(9-4x^2)
=∫(3/2)cosu du/ (3cosu)
=(1/2)u + C'
=(1/2)arcsin(2x/3) + C'
=(1/8)∫(-8x)/√(9-4x^2) dx +∫dx/√(9-4x^2)
=(1/4) √(9-4x^2) +∫dx/√(9-4x^2)
=(1/4) √(9-4x^2) +(1/2)arcsin(2x/3) + C
//
let
x=(3/2)sinu
dx=(3/2)cosu du
∫dx/√(9-4x^2)
=∫(3/2)cosu du/ (3cosu)
=(1/2)u + C'
=(1/2)arcsin(2x/3) + C'
本回答由网友推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
展开全部
∫[x³/(9+x²)]dx
=∫[x -9x/(9+x²)]dx
=∫xdx -(9/2)∫1/(x²+9)d(x²+9)
=½x² -(9/2)ln(x²+9) +C
=½[x²-9ln(x²+9)]+C
=∫[x -9x/(9+x²)]dx
=∫xdx -(9/2)∫1/(x²+9)d(x²+9)
=½x² -(9/2)ln(x²+9) +C
=½[x²-9ln(x²+9)]+C
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
