∫ (x^2+x^4)^1/2 dx
= ∫[x^2(x^2+x^4)]^1/2dx
= ∫x[(1+x^2)]^1/2dx 设x=tant,则dx=(sect)^2dt ∫(x^2+x^4)^1/2dx
= ∫[x^2(x^2+x^4)]^1/2dx
= ∫x[(1+x^2)]^1/2dx
= ∫tant{[1+(tant)^2]}^1/2* (sect)^2dt
= ∫tantsect* (sect)^2dt
= ∫(sect)^2d(secttant)
= (1/3)(sect)^3+C
由tant=x得到,sect=(x^2+1)^(1/2)
∫(x^2+x^4)^1/2dx =(1/3)(x^2+1)^(3/2)+C