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Integral of x/(1+x²)² from 1 to ∞

Short answer

Substitute t=1+x², dt=2x dx. The integral becomes (1/2)∫_{2}^{∞} t^{-2} dt = (1/2)[−1/t]_{2}^{∞} = (1/2)(0+1/2) = 1/4.

Evaluate ∫₁^∞ x dx/(1+x²)².

Let I = ∫₁^∞ x dx/(1 + x²)². This is an improper integral (infinite upper limit), defined as lim_(b→∞) ∫₁^b.

Put t = 1 + x², dt = 2x dx. Limits: x = 1 → t = 2; x = b → t = 1 + b².

I = lim_(b→∞) (1/2) ∫₂^(1+b²) t⁻² dt
  = lim (1/2) [−1/t]₂^(1+b²)
  = lim (1/2) [1/2 − 1/(1 + b²)]
  = (1/2)(1/2 − 0)

Answer: I = 1/4 (the integral converges).

✓ Answer checked by Claude Opus 5.5 (high effort)

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