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x>1 d(x^2 arctan根号下X-1)等于多少
人气:260 ℃ 时间:2020-02-05 10:26:45
解答
令根号下x-1=t,则x=t^2+1,t>0
d(x^2 arctan根号下X-1)=d((t^2+1)^2 arctant)=[2(t^2+1)*2t*arctant+t^2+1)^2*1/(t^2+1)]dt
=(t^2+1)*(1+4arctant)dt
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