求不定积分
晓*** 2007-12-02 21:31:43 举报
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令t=[(x-1)/(x+1)]^(1/3),则x=(1+t^3)/(1-t^3),原积分化为 ∫3/(2t^2) dt=-3/(2t)+C=-3/2×[(x+1)/(x-1)]^(1/3)+C
1/5÷2/3=1/5×3/2=3/10小时
(38+41)x2.5=197.5千米
连接OC;∵AB=4,O是AB中点,且△ABC是直角三角形,∴OC=2...