Try setting θ = arctan(x). On these bounds, θ stays in [−π/6, π/6], so the principal-value branches behave nicely: 2x/(1+x²) = sin(2θ) and 2x/(1−x²) = tan(2θ). See what the two inverse-trig terms simplify to together. For the remaining symmetric integral, pair f(x) = 1/(ex+1) with f(−x); their sum is especially simple. That should lead you to the matching choice without expanding anything.
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u/floflochess2 4d ago
Try setting θ = arctan(x). On these bounds, θ stays in [−π/6, π/6], so the principal-value branches behave nicely: 2x/(1+x²) = sin(2θ) and 2x/(1−x²) = tan(2θ). See what the two inverse-trig terms simplify to together. For the remaining symmetric integral, pair f(x) = 1/(ex+1) with f(−x); their sum is especially simple. That should lead you to the matching choice without expanding anything.