∫ x/√(x² + 4x + 5) dx
= (1/2) ∫ (2x + 4 - 4)/√(x² + 4x + 5) dx
= (1/2) ∫ (2x + 4)/√(x² + 4x + 5) dx - (1/2)(4) ∫ 1/√(x² + 4x + 5) dx
= (1/2) ∫ (x² + 4x + 5)^(-1/2) d(x² + 4x + 5) - 2 ∫ 1/√[(x + 2)² + 1] d(x + 2)
= √(x² + 4x + 5) - 2ln|x + 2 + √[(x + 2)² + 1]| + C