Evaluate: ∫x+2√x²+5x+6dx

Let I = ∫x+2√x²+5x+6dx
Put x+2 = λ(d/dx(x²+5x+6)) + μ
x+2 = 2λx + 5λ + μ
Comparing coefficients of x both sides
1 = 2λ ⇒ λ = 1/2
Comparing constant terms both sides,
2 = 5λ + μ
or, 2 = 5(1/2) + μ
μ = 2 - 5/2 = -1/2
Therefore, ∫x+2√x²+5x+6dx = ∫(1/2)(2x+5) - 1/2√x²+5x+6dx
(x+2 = λ(2x+5) + μ)
I = ∫(1/2)(2x+5)√x²+5x+6dx - 1/2∫dx√x²+5x+6
Therefore, I = I₁ - I₂ — (1)
I₁ = ∫(1/2)(2x+5)√x²+5x+6dx
Put x²+5x+6 = t ⇒ (2x+5)dx = dt
= 1/2∫dt√t = 1/2 ∫t^(1/2)dt = 1/2 [t^(3/2)/(3/2)] + C = t^(3/2)/3 + C = √t³/3 + C = √x²+5x+6³/3 + C
I₂ = 1/2∫dx√x²+5x+6
1/2∫dx√x²+5x+25/4 - 25/4 + 6 = 1/2∫dx√(x+5/2)² - (1/2)²
1/2 log|(x+5/2) + √(x+5/2)² - (1/2)²| + C
1/2 log|(x+5/2) + √x²+5x+6| + C
Substituting the values of I₁ and I₂ in (1),
I = √x²+5x+6³/3 + 1/2 log|(x+5/2) + √x²+5x+6| + C