If α = ∫₀¹ (e^(9x+3tan⁻¹x))(12+9x²)/(1+x²) dx, where tan⁻¹x takes only principal values, then the value of (logₑ|1+α|/3π/4) is

α = ∫₀¹ (e^(9x+3tan⁻¹x))(12+9x²)/(1+x²) dx
Let z = 9x + 3tan⁻¹x ⇒ dz = (12+9x²)/(1+x²) dx
∴ α = ∫₉⁺³π/₄⁰ e^z dz = e⁹⁺³π/₄ - 1 ⇒ logₑ(α+1) = 9 + 3π/4
⇒ logₑ|1+α|/3π/4 = 9