Đáp án:

\( f(x) + x(x+1)f'(x) = (x+1)^2 \)

\(\Leftrightarrow \frac{f(x)}{x+1} + x f'(x) = x+1\)

\( \left( \frac{x}{x+1} \right)' = \frac{1}{(x+1)^2} \)

\(\Leftrightarrow \frac{1}{(x+1)^2} \cdot f(x) + \frac{x}{x+1} \cdot f'(x) = 1 \)

\(\Leftrightarrow \left( \frac{x}{x+1} \cdot f(x) \right)' =1\)

\( \frac{x}{x+1} \cdot f(x) = x + C \)

\( f(1) = 2 \Rightarrow C = 0 \Rightarrow f(x) = x+1 \)

\(\Rightarrow f(2) = 3 \Rightarrow \boxed{A}\)

page46

Đáp án:

\( x(x+1)f'(x) - f(x) = x^2(2x+3) \)

\( \left( \frac{x+1}{x} \right)' = -\frac{1}{x^2} \)

\( \Rightarrow \left( \frac{x+1}{x} \right)f'(x) - \frac{1}{x^2}f(x) = 2x+3 \)

\( \Rightarrow \left( \frac{x+1}{x}f(x) \right)' = 2x+3 \)

\( \Rightarrow \frac{x+1}{x}f(x) = x^2 + 3x + C \)

\( f(1) = 2 \Rightarrow C = 0 \Rightarrow f(x) = \frac{x^3 + 3x^2}{x+1} \)

\( \Rightarrow f(2) = \frac{20}{3} \Rightarrow \boxed{C}\)

page47

Đáp án:

Xét \( \left( \frac{f(x)}{x} \right)' = \frac{x f'(x) - f(x)}{x^2} \) : Khó xoay số

\( \left( \frac{f(x)}{x^2} \right)' = \frac{x^2 f'(x) - 2x f(x)}{x^4} \) : Có thể xoay số

Nhân 2 vế với \( x\):  
\( x^2 f(x) + 2x f(x) = x^2 f'(x) - x^4 \)

\( \Leftrightarrow x^2 f'(x) + x^4 = x^2 f'(x) - 2x f(x) \)

\( \Leftrightarrow \frac{x^2 f'(x) - 2x f(x)}{x^4} = \frac{f(x)}{x^2} + 1 \)

\( \Rightarrow \left( \frac{f(x)}{x^2} \right)' = \frac{f(x)}{x^2} + 1 \)

Đặt \( h(x) = \frac{f(x)}{x^2} + 1 \Rightarrow  h'(x) = h(x) \),  
\( \frac{h'(x)}{h(x)}= 1 \Rightarrow \ln h(x) = x + C \)  

\( f(1) = e \Rightarrow h(1) = e+1 \Rightarrow \ln(e+1) = 1 + C \),  
\( \Rightarrow C = \ln(e+1) - 1 \)

\( \Rightarrow |h(x)| = e^{x + \ln(e+1) - 1} \Rightarrow (e+1)e^{x-1} \)

\(\Rightarrow \frac{f(x)}{x^2} + 1 = (e+1)e^{x-1} \)

\(\Rightarrow f(x) = x^2[(e+1)e^{x-1} - 1] \)

\(\Rightarrow f(2) = 4[(e+1)e - 1] = 4e^2 + 4e - 4 \Rightarrow \boxed{A}\)

page48

Đáp án:

\( (f(x))^3 + f(x) = x  \Rightarrow f'(x)(f(x))^3 + f'(x)f(x) = x f'(x) \)

\( \Rightarrow \int_0^2 f'(x)(f(x))^3 \, dx + \int_0^2 f'(x)f(x) \, dx = \int_0^2 x f'(x) \, dx \)

\( \Rightarrow \frac{(f(x))^4}{4} \big|_0^2 + \frac{(f(x))^2}{2} \Big|_0^2 = \int_0^2 x f'(x) \, dx \)

\( (f(0))^3 + f(0) = 0 \Rightarrow f(0)(1 + (f(0))^2) = 0 \Rightarrow f(0) = 0 \)

\( (f(2))^3 + f(2) - 2 = 0 \Rightarrow f(2) = 1 \)

\(\Rightarrow \int_0^2 x f'(x) \, dx = \frac{1}{4} + \frac{1}{2} = \frac{3}{4} \)

Đặt:  \( \begin{cases} 
u = x \\ 
dv = f'(x) \, dx 
\end{cases} 
\Rightarrow 
\begin{cases} 
du = dx \\ 
v = f(x) 
\end{cases}\)

\( \frac{3}{4} = \int_0^2 x f'(x) \, dx = x f(x)  \Big|_0^2 - \int_0^2 f(x) \, dx = 2 - \int_0^2 f(x) \, dx\)

\(\Rightarrow \int_0^2 f(x) \, dx = 2 - \frac{3}{4} = \frac{5}{4} \quad \Rightarrow \boxed{A}\)

page49

Đáp án:

\( \ln f(x) + \ln e^{f(x) - 1} = \ln \left[ (x^2 + 1)e^{x^2} \right] \)

\( \Rightarrow \ln f(x) \cdot e^{f(x) - 1} = \ln \left[ (x^2 + 1)e^{x^2} \right] \)

\( \Rightarrow f(x) \cdot e^{f(x) - 1} = (x^2 + 1)e^{x^2} \quad (1) \)

\( g(t) = t \cdot e^{t - 1} \)  đồng biến trên \( (0, +\infty) \).

(1) \( \Leftrightarrow g(f(x)) = g(x^2 + 1) \Leftrightarrow f(x) = x^2 + 1 \).

 \( I = \int_0^1 x f(x) \, dx = \int_0^1 (x^3 + x) \, dx = \frac{3}{4} \Rightarrow \boxed{D} \)

page50