Question

如图，高等数学求不定积分

如图，高等数学求不定积分求不定积分，急

Answer 1

I = (1/2)∫<√2, +∞> dx^2/[x^2√(x^4+x^2-1)]
= (1/2)∫<2, +∞> du/[u√(u^2+u-1)]
= (1/2)∫<2, +∞> du/{u√[(u+1/2)^2-5/4)]}
令 u+1/2 = (√5/2)secv 则 u =(√5/2)secv-1/2
du =(√5/2)secvtanvdv
I = (1/2)∫<arccos(1/√5), π/2> secvdv/[(√5/2)secv-1/2]
= ∫<arccos(1/√5), π/2> dv/(√5-cosv) 半角代换
= [arctan{2tan(v/2)/(√5-1)}] <arccos(1/√5), π/2>
= arctan(√5+1) - π/4

追问

最后答案不一样，前面过程不错