Đáp án:
$\begin{array}{l}
a)I = \int {\left( {{x^2} + 1} \right)\sin 2xdx} \\
Dat:\left\{ \begin{array}{l}
{x^2} + 1 = u \Rightarrow du = 2xdx\\
\sin 2xdx = dv \Rightarrow v = \dfrac{{ - 1}}{2}.cos2x
\end{array} \right.\\
\Rightarrow I = \left( {{x^2} + 1} \right).\left( {\dfrac{{ - 1}}{2}.\cos 2x} \right) + \int {\dfrac{1}{2}.\cos 2x.2xdx} \\
= - \dfrac{1}{2}.\left( {{x^2} + 1} \right).cos2x + \int {\cos 2x.x.dx} \\
= - \dfrac{1}{2}.\left( {{x^2} + 1} \right).\cos 2x + x.\dfrac{1}{2}.\sin 2x - \int {\dfrac{1}{2}.\sin 2xdx} \\
= - \dfrac{1}{2}.\left( {{x^2} + 1} \right).\cos 2x + \dfrac{1}{2}.x.\sin 2x + \dfrac{1}{2}.\dfrac{1}{2}.\cos 2x + C\\
= - \dfrac{1}{2}.\left( {{x^2} + 1} \right).\cos 2x + \dfrac{1}{2}.x.\sin 2x + \dfrac{1}{4}.\cos 2x + C\\
= \dfrac{1}{4}.\left( { - 2{x^2} - 1} \right).\cos 2x + \dfrac{1}{2}x.\sin 2x + C\\
b)\int {\dfrac{{{x^2} + 3x - 3}}{{x + 2}}dx} \\
= \int {\dfrac{{{x^2} + 4x + 4 - x - 2 - 5}}{{x + 2}}dx} \\
= \int {x + 2 - 1 - \dfrac{5}{{x + 2}}dx} x\\
= \int {x + 1 - \dfrac{5}{{x + 2}}dx} \\
= \dfrac{1}{2}{x^2} + x - 5\ln \left| {x + 2} \right| + C
\end{array}$
