求limx->正无穷∫(0,x)(arctant)^2dt/根号下(x^2+1),
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求limx->正无穷∫(0,x)(arctant)^2dt/根号下(x^2+1),
求limx->正无穷∫(0,x)(arctant)^2dt/根号下(x^2+1),
求limx->正无穷∫(0,x)(arctant)^2dt/根号下(x^2+1),
用罗必达法则,limx->正无穷∫(0,x)(arctant)^2dt/根号下(x^2+1)=limx->正无穷 (arctanx)^2 × 根号下(x^2+1)/x
limx->正无穷 (arctanx)^2 =\pi^2/4
limx->正无穷 根号下(x^2+1) /x=1
所以原极限=\pi^2/4(圆周率平方除以4)
你没有发现arctan(t)求导后是分母???这个意思就是告诉你,把分母放进微分后:d(arctan(t)),这下就简单了吧,求出积分,再来个极限!