F(x)=∫(x²,0) dt/[(1+5t²)√(1+t²)]F'(x)=2x*1/[(1+5x^4)√(1+x^4)]当x＞0时，F'(x)＞0，F(x)单调递增当x＜0时，F'(x)＜0，F(x)单调递减所以驻点x=0，F(0)=0为极小值点故F(x)≥0
