First, rewriting a bit,
[tex]\dfrac{x^3+4x^2+1}{(i+1)x+2} = \dfrac1{i+1} \cdot \dfrac{x^3+4x^2+1}{x+\frac2{i+1}}[/tex]
By the remainder theorem, the remainder upon dividing [tex]x^3+4x^2+1[/tex] by [tex]x+\frac2{i+1}[/tex] is equal to
[tex]\left(-\dfrac2{i+1}\right)^3 + 4\left(-\dfrac2{i+1}\right)^2 + 1 = \boxed{3 - 6i}[/tex]
which is to say,
[tex]\dfrac1{i+1} \cdot \dfrac{x^3+4x^2+1}{x+\frac2{i+1}} = q(x) + \dfrac{3-6i}{(i+1)\left(x+\frac2{i+1}\right)}= q(x) + \dfrac{3-6i}{(i+1)x+2}[/tex]
