f(x)=arctanx+C
令y=arctanx；则x=tany
因为
f'(x)=(arctanx)'+0
=1/(tany)'
=1/(siny/cosy)'
=1/[(cos^2y+sin^2y)/cos^2y]
=1/(1+tan^2y)
=1/(1+x^2)
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