第三题的2小问，怎么做?
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3.(2)设u=arctan(x^2),则x^2=tanu,2xdx=(secu)^2du,
∫x^3dx/[√(x^4+1)+1]
=(1/2)∫tanu(secu)^2du/(secu+1)
=(1/2)∫sinudu/[(1+cosu)(cosu)^2]
=(-1/2)∫dv/[(1+v)v^2](v=cosu)
=(-1/2)∫[1/(1+v)-1/v+1/v^2]dv
=(-1/2)[ln|(1+v)/v|-1/v]+c
=(-1/2)[ln|secu+1|-secu]+c
=(-1/2{ln|sec[arctan(x^2)+1|-sec[arctan(x^2)]}+c
=(-1/2)[ln|1/√(x^4+1)+1|-1/√(x^4+1)]+c.
∫x^3dx/[√(x^4+1)+1]
=(1/2)∫tanu(secu)^2du/(secu+1)
=(1/2)∫sinudu/[(1+cosu)(cosu)^2]
=(-1/2)∫dv/[(1+v)v^2](v=cosu)
=(-1/2)∫[1/(1+v)-1/v+1/v^2]dv
=(-1/2)[ln|(1+v)/v|-1/v]+c
=(-1/2)[ln|secu+1|-secu]+c
=(-1/2{ln|sec[arctan(x^2)+1|-sec[arctan(x^2)]}+c
=(-1/2)[ln|1/√(x^4+1)+1|-1/√(x^4+1)]+c.
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