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

Biết $\Large J=\int\limits_{1}^{4}{x{{\log }_{2}}xdx=16-\dfrac{a}{b\ln 2}}$ với $\Large a,b\in {{N}^{*}}$ ; $\Large \dfrac{a}{b}$ là phân số tối giản . Tính $\Large T=a+b$.

• A.

  A. $\Large T=11$

• B.

  B. $\Large T=19$

• C.

  C. $\Large T=13$

• D.

  D. $\Large T=17$

Tính $\Large J=\int\limits_{1}^{4}{x{{\log }_{2}}xdx}$

Đặt $\Large \left\{ \begin{align}  & u={{\log }_{2}}x \\  & dv=xdx \\ \end{align} \right.$ $\Large \Rightarrow \left\{ \begin{align}  & du=\dfrac{1}{x\ln 2}dx \\  & v=\dfrac{{{x}^{2}}}{2} \\ \end{align} \right.$

$\Large J=\int\limits_{1}^{4}{x{{\log }_{2}}xdx} =\left( \dfrac{{{x}^{2}}}{2}{{\log }_{2}}x \right)\left| \begin{align}  & 4 \\  & 1 \\ \end{align} \right.-\int\limits_{1}^{4}{\dfrac{1}{x\ln 2}.\dfrac{{{x}^{2}}}{2}dx}$

$\Large =16-\dfrac{1}{2\ln 2}\int\limits_{1}^{4}{xdx} =16-\dfrac{1}{2\ln 2}.\dfrac{{{x}^{2}}}{2}\left| \begin{align}  & 4 \\  & 1 \\ \end{align} \right.$

$\Large =16-\dfrac{1}{2\ln 2}.\dfrac{15}{2}=16-\dfrac{15}{4\ln 2}\Rightarrow a=15;b=4$

Vậy $\Large T=a+b=19$

Chọn đáp án B