Đáp án:
$\begin{array}{l}
a)\int {\sin \left( {\dfrac{x}{2}} \right).\cos \left( {\dfrac{x}{2}} \right)dx} \\
= \int {\dfrac{1}{2}.2.\sin \left( {\dfrac{x}{2}} \right).\cos \left( {\dfrac{x}{2}} \right)dx} \\
= \dfrac{1}{2}.\int {\sin \left( {2.\dfrac{x}{2}} \right)dx} \\
= \dfrac{1}{2}.\int {\sin xdx} \\
= \dfrac{1}{2}.\left( { - \cos x} \right) + C\\
= - \dfrac{1}{2}.\cos x + C\\
b)\int {\left( {2x - 1} \right).\ln xdx} \\
= \int {2x.\ln xdx} - \int {\ln xdx} \\
= 2.\int {x.\ln xdx} - \left( {x.\ln x - \int {x.\dfrac{1}{x}dx} } \right)\\
= 2.\left( {\dfrac{{{x^2}}}{2}.\ln x - \int {\dfrac{{{x^2}}}{2}.\dfrac{1}{x}dx} } \right) - x.\ln x + \int {dx} \\
= {x^2}.\ln x - \int {xdx} - x.\ln x + x\\
= {x^2}.\ln x - \dfrac{{{x^2}}}{2} - x.\ln x + x + C\\
= {x^2}.\ln x - x.\ln x - \dfrac{{{x^2}}}{2} + x + C
\end{array}$
