Biết $\int\limits_{1}^{e}{\dfrac{1-\ln x}{{{(x+\ln x)}^{2}}}\text{d}x=\dfrac{1}{ae+b}}$ với $a,b\in Z.$ Tính $T=2a+{{b}^{2}}$

Ta có: $\int\limits_{1}^{e}{\dfrac{1-\ln x}{{{(x+\ln x)}^{2}}}\text{d}x}$ $=\int\limits_{1}^{e}{\dfrac{\dfrac{1}{{{x}^{2}}}-\dfrac{\ln x}{{{x}^{2}}}}{{{(1+\dfrac{\ln x}{x})}^{2}}}\text{d}x}$ $=\int\limits_{1}^{e}{\dfrac{\text{d}\left( 1+\dfrac{\ln x}{x} \right)}{{{(1+\dfrac{\ln x}{x})}^{2}}}}$ $=\left. -\dfrac{1}{1+\dfrac{\ln x}{x}} \right|_{1}^{e}$ $=\dfrac{1}{e+1}$
$\Rightarrow \left\{ \begin{aligned}
& a=1 \\
& b=1 \\
\end{aligned} \right. $.
Khi đó: $ T=2a+{{b}^{2}}=3$.