Hướng dẫn giải

Ta có

\[I = \int\limits_1^2 {{{\left( {x + 1} \right)}^2}{e^{x - \frac{1}{x}}}dx = \int\limits_1^2 {\left( {{x^2} + 2x + 1} \right){e^{x - \frac{1}{x}}}} dx = } \int\limits_1^2 {\left( {{x^2} + 1} \right)} {e^{x - \frac{1}{x}}}dx + \int\limits_1^2 {2x{e^{x - \frac{1}{x}}}} dx.\]

Xét \({I_1} = \int\limits_1^2 {\left( {{x^2} + 1} \right)} {e^{x - \frac{1}{x}}}dx = \int\limits_1^2 {{x^2}.{e^{x - \frac{1}{x}}}} .\frac{{{x^2} + 1}}{{{x^2}}}dx = \int\limits_1^2 {{x^2}.{e^{x - \frac{1}{x}}}} d\left( {x - \frac{1}{x}} \right) = \int\limits_1^2 {{x^2}d\left( {{e^{x - \frac{1}{x}}}} \right)} \)

\( = {x^2}{e^{x - \frac{1}{x}}}\left| {_{\scriptstyle\atop\scriptstyle1}^{\scriptstyle2\atop\scriptstyle}} \right. - \int\limits_1^2 {{e^{x - \frac{1}{x}}}} d\left( {{x^2}} \right) = {x^2}{e^{x - \frac{1}{x}}}\left| {_{\scriptstyle\atop\scriptstyle1}^{\scriptstyle2\atop\scriptstyle}} \right. - \int\limits_1^2 {2x{e^{x - \frac{1}{x}}}} dx\)

\( \Rightarrow {I_1} + \int\limits_1^2 {2x{e^{x - \frac{1}{x}}}} dx = {x^2}{e^{x - \frac{1}{x}}}\left| {_{\scriptstyle\atop\scriptstyle1}^{\scriptstyle2\atop\scriptstyle}} \right. \Rightarrow I = {x^2}{e^{x - \frac{1}{x}}}\left| {_{\scriptstyle\atop\scriptstyle1}^{\scriptstyle2\atop\scriptstyle}} \right. = 4{e^{\frac{3}{2}}} - 1\)

\( \Rightarrow m = 4,n = 1,p = 3,q = 2.\)

Khi đó \(T = m + n + p + q = 4 + 1 + 3 + 2 = 10.\)

Chọn B.