If ∫21 dx/(x² - 2x + 4)^(3/2) = k/(k+5), then k is equal to

∫₂¹ dx/(x² - 2x + 4)^(3/2) = ∫₂¹ dx/[(x - 1)² + 3]^(3/2)
Substitute x - 1 = √3 tan θ, when x = 1 ⇒ θ = 0, when x = 2, ⇒ θ = tan⁻¹(1/√3) = π/6
dx = √3 sec²θ dθ
L = ∫π/6₀ √3 sec²θ dθ/[3 tan²θ + 3]^(3/2) = ∫π/6₀ √3 sec²θ dθ/[3 sec²θ]^(3/2) = ∫π/6₀ √3 sec²θ dθ/3√3 sec³θ = (1/3)∫π/6₀ cos θ dθ = (1/3) sin θ |π/6₀ = (1/3)[sin(π/6) - sin(0)] = (1/3)[1/2 - 0] = 1/6
1/6 = k/(k+5)
k + 5 = 6k
5 = 5k
k = 1