\(I = \int_0^2 {\left( {2x + \ln \left( {x + 1} \right)} \right)dx} = A + B\)

Tính \(A = \int_0^2 {2xdx} = \left. {{x^2}} \right|_0^2 = 4\)

Tính \(B = \int_0^2 {\left( {\ln \left( {x + 1} \right)} \right)} dx\)

Đặt \(\left\{ \begin{array}{l} u = \ln \left( {x + 1} \right)\\ dv = dx \end{array} \right.\) 

\(\Rightarrow \left\{ \begin{array}{l} du = \frac{{dx}}{{x + 1}}\\ v = x + 1 \end{array} \right.\)

Dùng công thức tích phân từng phần:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
B = \int_0^2 {\left( {\ln \left( {x + 1} \right)} \right)dx} \\
 = \left. {\left( {x + 1} \right).\ln \left( {x + 1} \right)} \right|_0^2 - \int_0^2 {\frac{{x + 1}}{{x + 1}}dx} 
\end{array}\\
{ = \left. {3\ln 3 - x} \right|_0^2 = 3\ln 3 - 2}
\end{array}\)

Vậy: \(I = \int_0^2 {\left( {2x + \ln \left( {x + 1} \right)} \right)} dx = 3\ln 3 + 2\)