∫1/（x∧4*√（x∧2+1））dx

网友回答

原式=∫(sect)^2dt/((tant)^4sect) (令x=tant,则sint=x/√(x^2+1))
=∫(1-(sint)^2)d(sint)/(sint)^4
=1/sint-(1/3)/(sint)^3+C (C是积分常数)
=√(x^2+1))/x-(1/3)(√(x^2+1))/x)^3+C.