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Question:

If 2∫₁⁰tan⁻¹xdx = ∫₁⁰cot⁻¹(1-x+x²)dx, then ∫₁⁰tan⁻¹(1-x+x²)dx is equal to:

log2

π/2+log2

log4

π/2-log4

Solution:

We know that tan⁻¹x + cot⁻¹x = π/2
2∫₁⁰tan⁻¹xdx = ∫₁⁰cot⁻¹(1-x+x²)dx, then
2∫₁⁰tan⁻¹xdx = ∫₁⁰[π/2 - tan⁻¹(1-x+x²)]dx
If I = ∫₁⁰tan⁻¹(1-x+x²)dx = π/2 - ∫₁⁰tan⁻¹xdx
Integrating ∫₁⁰tan⁻¹xdx
I = ∫₁⁰tan⁻¹xdx = [x(tan⁻¹x)]₁⁰ - ∫₁⁰ x/(1+x²)dx = π/4 - [1/2 ln(1+x²)]₁⁰ = π/4 - (1/2)ln2
If I = π/2 - [π/4 - (1/2)ln2] = π/4 + (1/2)ln2 = π/4 + ln√2