Đáp án:

\(x(x+1)f'(x) + f(x) = x(x+1) \quad \text{Chia 2 vế cho } (x+1)^2\)

\(\Rightarrow \frac{x}{x+1}f'(x) + \frac{f(x)}{(x+1)^2} = \frac{x}{x+1}\) (Gợi ý: \( \left[\frac{x}{x+1}\right]' = (1 - \frac{1}{(x+1)})' =\frac{1}{(x+1)^2}\) )

Vì: \(\left[\frac{x}{x+1}f(x)\right]' = \frac{x}{x+1}f'(x) + \frac{f(x)}{(x+1)^2}\)

\(\Rightarrow \left[\frac{x}{x+1}f(x)\right]' = \frac{x}{x+1}.\)

\(\Rightarrow \frac{x}{x+1}f(x) = \int \frac{x}{x+1} dx = x - \ln{|x+1|} + C.\)

\(f(1) = -2\ln{2} \quad \Rightarrow -\ln{2}  = 1-\ln{2} +C\quad \Rightarrow C = -1.\)

\(\Rightarrow \frac{x}{x+1}f(x) =  - \ln{|x+1|} +x.\)

\(\Rightarrow \frac{2}{3} f(2) = -\ln{3} + 1 \quad \Rightarrow f(2) = \frac{3}{2} - \frac{3}{2}\ln{2}.\)

\(\Rightarrow a^2 + b^2 = \frac{9}{2} \quad \Rightarrow \boxed{\text{B}}.\)

Gợi ý: Vì \(\left[\frac{x}{x+1}f(x)\right]' = \frac{1}{(x+1)^2}f(x) + \frac{x}{x+1}f'(x).\)