Lời giải:Đặt \(\left\{ \begin{array}{l}
u = \ln \left( {9 - {x^2}} \right)\\
dv = dx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = \frac{{ - 2x}}{{9 - {x^2}}}dx\\
v = x + 3
\end{array} \right..\)
Khi đó \(I = \left( {x + 3} \right)\ln \left( {9 - {x^2}} \right)\left| {\begin{array}{*{20}{c}}
2\\
1
\end{array}} \right. + 2\int\limits_1^2 {\frac{{x\left( {x + 3} \right)}}{{9 - {x^2}}}{\rm{d}}x} = 5\ln 5 - 4\ln 8 + 2\int\limits_1^2 {\left( { - 1 + \frac{3}{{3 - x}}} \right){\rm{d}}x} \)
\( = 5\ln 5 - 12\ln 2 - 2\left( {x + 3\ln \left| {3 - x} \right|} \right)\left| {\begin{array}{{20}{c}}
2\\
1
\end{array}} \right. = 5\ln 5 - 6\ln 2 - 2 \to \left\{ \begin{array}{l}
a = 5\\
b = - 6\\
c = - 2
\end{array} \right. \to P = 13.\)
