∫[（x＾3+1）/（x＾2+1）＾]2dx不定积分?

Question

∫[（x＾3+1）/（x＾2+1）＾]2dx 不定积分?

Answer 1

∫(x³+1)/(x²+1)² dx
= ∫(1-x)/(x²+1)² dx + ∫x/(x²+1) dx
= J + (1/2)ln(x²+1)
令x=tany,dx=sec²y dy,siny=x/√(x²+1),cosy=1/√(x²+1)
J = ∫(1-tany)/sec⁴y * sec²y dy
= ∫(1-tany)cos²y dy
= ∫cos²y dy - ∫sinycosy dy
= (1/2)∫(1+cos2y) - (1/2)∫sin2y dy
= y/2 + 1/4*sin2y + 1/4*cos2y
= (1/2)arctanx + (1/2)*x/(x²+1) + 1/4*[2/(x²+1)-1]
= (1/2)arctanx + x/[2(x²+1)] + (1-x²)/[2(x²+1)]
原积分= (1/2)arctanx + x/[2(x²+1)] + (1-x²)/[2(x²+1)] + (1/2)ln(x²+1) + C
= (1/2)[(x+1)/(x²+1) + ln(x²+1) + arctanx] + C