Hướng dẫn giải

Xét $I = \int\limits_0^2 {2x\ln \left( {1 + x} \right)dx.}$

Đặt $\left\{ \begin{array}{l}u = \ln \left( {1 + x} \right)\\dv = 2xdx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \frac{1}{{1 + x}}dx\\v = {x^2} - 1\end{array} \right.$ .

Ta có $I = \left( {{x^2} - 1} \right)\ln \left( {x + 1} \right)\left| {_{\scriptstyle\atop\scriptstyle0}^{\scriptstyle2\atop\scriptstyle}} \right. - \int\limits_0^2 {\frac{{{x^2} - 1}}{{x + 1}}dx}$

$= 3\ln 3 - \int\limits_0^2 {\left( {x - 1} \right)dx} = 3\ln 3 - \left( {\frac{{{x^2}}}{2} - x} \right)\left| {_{\scriptstyle\atop\scriptstyle0}^{\scriptstyle2\atop\scriptstyle}} \right. = 3\ln 3.$

Vậy $a = 3,b = 3 \Rightarrow 3a + 4b = 21.$