Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\,\,\,\,\,{I_1} = \int {x.\sin xdx} \\
\left\{ \begin{array}{l}
u = x\\
v' = \sin x
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = 1\\
v =  - \cos x
\end{array} \right.\\
 \Rightarrow {I_1} = x.\left( { - \cos x} \right) - \int {1.\left( { - \cos x} \right)dx}  =  - x.\cos x + \int {\cos xdx} \\
 =  - x.\cos x + \sin x + C\\
2,\\
{I_2} = \int {x.\cos xdx} \\
\left\{ \begin{array}{l}
u = x\\
v' = \cos x
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
u' = 1\\
v = \sin x
\end{array} \right.\\
 \Rightarrow {I_2} = x.\sin x - \int {1.\sin xdx}  = x.\sin x - \left( { - \cos x} \right) + C = x.\sin x + \cos x + C\\
3,\\
{I_3} = \int {\left( {{x^2} + 5} \right).\sin xdx} \\
\left\{ \begin{array}{l}
u = {x^2} + 5\\
v' = \sin x
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = 2x\\
v =  - \cos x
\end{array} \right.\\
{I_3} = \left( {{x^2} + 5} \right).\left( { - \cos x} \right) - \int {2x.\left( { - \cos x} \right)dx} \\
 =  - \left( {{x^2} + 5} \right).\cos x + 2\int {x\cos xdx} \\
 =  - \left( {{x^2} + 5} \right).\cos x + 2.{I_2}\\
 =  - \left( {{x^2} + 5} \right).\cos x + 2.\left( {x.\sin x + \cos x} \right) + C\\
 =  - \left( {{x^2} + 3} \right).\cos x + 2x\sin x + C\\
4,\\
{I_4} = \int {\left( {{x^2} + 2x + 3} \right).\cos xdx} \\
\left\{ \begin{array}{l}
u = {x^2} + 2x + 3\\
v' = \cos x
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = 2x + 2\\
v = \sin x
\end{array} \right.\\
 \Rightarrow {I_4} = \left( {{x^2} + 2x + 3} \right).\sin x - \int {\left( {2x + 2} \right).\sin xdx} \\
 = \left( {{x^2} + 2x + 3} \right).\sin x - 2\int {x\sin xdx}  - 2\int {\sin xdx} \\
 = \left( {{x^2} + 2x + 3} \right).\sin x - 2.{I_1} + 2\cos x + C\\
 = \left( {{x^2} + 2x + 3} \right).\sin x - 2.\left( { - x.\cos x + \sin x} \right) + 2\cos x + C\\
 = \left( {{x^2} + 2x + 1} \right).\sin x + 2x\cos x + 2\cos x + C\\
5,\\
{I_5} = \int {x.\sin 2xdx} \\
\left\{ \begin{array}{l}
u = x\\
v' = \sin 2x
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = 1\\
v =  - \frac{1}{2}\cos 2x
\end{array} \right.\\
 \Rightarrow {I_5} = x.\left( { - \frac{1}{2}\cos 2x} \right) - \int {1.\frac{{ - 1}}{2}\cos 2xdx} \\
 =  - \frac{1}{2}x.\cos 2x + \frac{1}{2}\int {\cos 2xdx} \\
 =  - \frac{1}{2}x.\cos 2x + \frac{1}{4}\sin 2x + C\\
6,\\
{I_6} = \int {x.\cos 2xdx} \\
\left\{ \begin{array}{l}
u = x\\
v' = \cos 2x
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = 1\\
v = \frac{1}{2}\sin 2x
\end{array} \right.\\
 \Rightarrow {I_6} = x.\frac{1}{2}\sin 2x - \int {1.\frac{1}{2}\sin 2xdx} \\
 = \frac{1}{2}.x.\sin 2x - \frac{1}{2}\int {\sin 2xdx} \\
 = \frac{1}{2}.x.\sin 2x + \frac{1}{4}.\cos 2x + C\\
7,\\
{I_7} = \int {x.{e^x}dx} \\
\left\{ \begin{array}{l}
u = x\\
v' = {e^x}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = 1\\
v = {e^x}
\end{array} \right.\\
 \Rightarrow {I_7} = x.{e^x} - \int {1.{e^x}dx}  = x.{e^x} - {e^x} + C\\
8,\\
{I_8} = \int {\ln xdx} \\
\left\{ \begin{array}{l}
u = \ln x\\
v' = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
u' = \frac{1}{x}\\
v = x
\end{array} \right.\\
 \Rightarrow {I_8} = x\ln x - \int {\frac{1}{x}.x.dx}  = x.\ln x - x + C
\end{array}\)