Biết $I=\int\limits_{1}^{4}{x{{\log }_{2}}xdx}=16-\dfrac{a}{b\ln...

Ta có $I=\int\limits_{1}^{4}{x{{\log }_{2}}xdx}=\dfrac{1}{\ln 2}\int\limits_{1}^{4}{x\ln xdx}$. Đặt $\left\{ \begin{aligned}
& u=\ln x \\
& dv=xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{x}dx \\
& v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.$.
Lúc đó $I=\dfrac{1}{\ln 2}\left( \dfrac{{{x}^{2}}}{2}\ln x\left| \begin{aligned}
& ^{4} \\
& _{1} \\
\end{aligned} \right.-\dfrac{1}{2}\int\limits_{1}^{4}{xdx} \right)=\dfrac{1}{\ln 2}\left( \dfrac{{{x}^{2}}}{2}\ln x\left| \begin{aligned}
& ^{4} \\
& _{1} \\
\end{aligned} \right.-\dfrac{{{x}^{2}}}{4}\left| \begin{aligned}
& ^{4} \\
& _{1} \\
\end{aligned} \right. \right)$
$=\dfrac{1}{\ln 2}\left( 8\ln 4-\dfrac{15}{4} \right)=\dfrac{1}{\ln 2}\left( 16\ln 2-\dfrac{15}{4} \right)=16-\dfrac{15}{4\ln 2}$
Vậy $\left\{ \begin{aligned}
& a=15 \\
& b=4 \\
\end{aligned} \right. $ nên $ T=a+b=15+4=19$.