Question

高数，求积分，如图

Answer 1

let
u = arctanx
du = dx/(1+x^2)
tanu = x
1/x = 1/tanu
= tan(π/2-u)
tan(1/x) = π/2-u
ln[tan(1/x)] = ln(π/2-u)
∫ ln[tan(1/x)] / ( 1+x^2) dx

=∫ ln(π/2-u) du
=uln(π/2-u) + ∫ u/(π/2-u) du
=uln(π/2-u) + ∫ [-1 + (π/2)/(π/2-u) ] du
=uln(π/2-u) -u - (π/2)ln|π/2-u| +C
=arctanx. ln|arctan(1/x)| - arctanx -(π/2)ln|arctan(1/x)| +C

追问

tan(1/x) = π/2-u
这一步怎么得到的？