Cho hàm số $y=f\left(x\right)$ liên tục trên $\mathbb{R}\setminus\left\{0;-1\right\}$ thỏa mãn điều kiện $f\left(1\right)=-2\ln2$ và $x\left(x+1\right)\cdot f'\left(x\right)+f\left(x\right)=x^2+x$. Giá trị $f\left(2\right)=a+b\ln3$, với $a,\,b\in\mathbb{Q}$. Tính $a^2+b^2$.
Nhận xét: $\left(\dfrac{x}{x+1}\right)^\prime=\dfrac{1}{(x+1)^2}$.
\begin{eqnarray*}
&x\left(x+1\right)\cdot f'\left(x\right)+f\left(x\right)&=x^2+x\\
\Leftrightarrow&x\left(x+1\right)\cdot f'\left(x\right)+f\left(x\right)&=x(x+1)\\
\Leftrightarrow&\dfrac{x}{x+1}\cdot f'\left(x\right)+\dfrac{1}{(x+1)^2}\cdot f\left(x\right)&=\dfrac{x}{x+1}\\
\Leftrightarrow&\dfrac{x}{x+1}\cdot f'\left(x\right)+\left(\dfrac{x}{x+1}\right)^\prime\cdot f\left(x\right)&=\dfrac{x}{x+1}\\
\Leftrightarrow&\left[\dfrac{x}{x+1}\cdot f\left(x\right)\right]^\prime&=\dfrac{x}{x+1}\\
\Leftrightarrow&\displaystyle\int\left[\dfrac{x}{x+1}\cdot f\left(x\right)\right]^\prime\mathrm{d}x&=\displaystyle\int\dfrac{x}{x+1}\mathrm{d}x\\
\Leftrightarrow&\dfrac{x}{x+1}\cdot f\left(x\right)&=\displaystyle\int\left(1-\dfrac{1}{x+1}\right)\mathrm{d}x\\
\Leftrightarrow&\dfrac{x}{x+1}\cdot f\left(x\right)&=x-\ln|x+1|+C.
\end{eqnarray*}
Với $x=1$ ta có $$\dfrac{1}{2}\cdot\left(-2\ln2\right)=1-\ln2+C\Leftrightarrow C=-1.$$
Với $x=2$ ta có $$\dfrac{2}{3}\cdot f\left(2\right)=2-\ln3-1\Leftrightarrow f(2)=\dfrac{3}{2}-\dfrac{3}{2}\ln3$$
Suy ra $a=\dfrac{3}{2}$ và $b=-\dfrac{3}{2}$.
Vậy $a^2+b^2=\dfrac{9}{4}+\dfrac{9}{4}=\dfrac{9}{2}$.