Phương pháp giải:

Sử dụng công thức tích phân từng phần: \(\int\limits_a^b {udv}  = \left. {uv} \right|_a^b - \int\limits_a^b {vdu} \).

Lời giải chi tiết:

Ta có: \(\left( {\dfrac{1}{{{x^2} + 1}}} \right)' = \dfrac{{ - 2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}\).

\(\begin{array}{l}\int\limits_1^2 {\dfrac{{x\ln xdx}}{{{{\left( {{x^2} + 1} \right)}^2}}}}  =  - \dfrac{1}{2}\int\limits_1^2 {\dfrac{{ - 2x\ln x}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx}  =  - \dfrac{1}{2}\int\limits_1^2 {\ln x\left( {\dfrac{1}{{{x^2} + 1}}} \right)'dx}  =  - \dfrac{1}{2}\int\limits_1^2 {\ln xd\left( {\dfrac{1}{{{x^2} + 1}}} \right)} \\ =  - \dfrac{1}{2}\left( {\left. {\ln x.\dfrac{1}{{{x^2} + 1}}} \right|_1^2 - \int\limits_1^2 {\dfrac{{dx}}{{x\left( {{x^2} + 1} \right)}}} } \right) =  - \dfrac{1}{2}\left( {\dfrac{{\ln 2}}{5} - I} \right)\end{array}\)

\(I = \int\limits_1^2 {\dfrac{{dx}}{{x\left( {{x^2} + 1} \right)}}}  = \int\limits_1^2 {\dfrac{{xdx}}{{{x^2}\left( {{x^2} + 1} \right)}}} \)

Đặt \(t = {x^2} + 1 \Leftrightarrow dt = 2xdx \Leftrightarrow xdx = \dfrac{{dt}}{2}\).

Đổi cận \(\left\{ \begin{array}{l}x = 1 \Leftrightarrow t = 2\\x = 2 \Leftrightarrow t = 5\end{array} \right.\)

\(\begin{array}{l} \Rightarrow I = \dfrac{1}{2}\int\limits_2^5 {\dfrac{{dt}}{{\left( {t - 1} \right)t}}}  = \dfrac{1}{2}.\int\limits_2^5 {\left( {\dfrac{1}{{t - 1}} - \dfrac{1}{t}} \right)dt} \\ = \dfrac{1}{2}\left. {\left( {\ln \left| {t - 1} \right| - \ln \left| t \right|} \right)} \right|_2^5 = \dfrac{1}{2}\left( {\ln 4 - \ln 5 - ln1 + ln2} \right) = \dfrac{1}{2}\left( {3\ln 2 - \ln 5} \right)\end{array}\)

\(\begin{array}{l} \Rightarrow \int\limits_1^2 {\dfrac{{x\ln xdx}}{{{{\left( {{x^2} + 1} \right)}^2}}}}  =  - \dfrac{1}{2}\left( {\dfrac{{\ln 2}}{5} - \dfrac{1}{2}\left( {3\ln 2 - \ln 5} \right)} \right) = \dfrac{{13}}{{20}}\ln 2 - \dfrac{1}{4}\ln 5\\ \Rightarrow a = \dfrac{{13}}{{20}};\,\,b = 0;\,\,c =  - \dfrac{1}{4} \Rightarrow a + b + c = \dfrac{2}{5}\end{array}\)

Chọn B.