Cho hàm số $f\left( x \right)=\left\{ \begin{aligned} &...

Hàm số $f\left( x \right)$ có đạo hàm tại ${{x}_{0}}=0$ khi và chỉ khi:
$\left\{ \begin{aligned}
& \underset{x\to {{0}^{+}}}{\mathop{\lim }} f\left( x \right)=\underset{x\to {{0}^{-}}}{\mathop{\lim }} f\left( x \right) \\
& f'\left( {{0}^{+}} \right)=f'\left( {{0}^{-}} \right) \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 1+a=0 \\
& 1=b \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-1 \\
& b=1 \\
\end{aligned} \right.$
Khi đó $f\left( x \right)=\left\{ \begin{aligned}
& {{e}^{x}}-1 \text{khi} x\ge 0 \\
& -{{x}^{3}}+x \text{khi} x<0 \\
\end{aligned} \right. $nên $ I=\int\limits_{\ln \left( \dfrac{e}{e+1} \right)}^{-\ln \left( e+1 \right)}{\dfrac{1}{1-{{e}^{x}}}f\left( \ln \left( {{e}^{-x}}-1 \right) \right)dx}$.
Đặt $t=\ln \left( {{e}^{-x}}-1 \right)\Rightarrow dt=\dfrac{-{{e}^{-x}}}{{{e}^{-x}}-1}dx=-\dfrac{1}{1-{{e}^{x}}}dx\Rightarrow -dt=\dfrac{1}{1-{{e}^{x}}}dx$
Đổi biến:
Với $x=\ln \dfrac{e}{e+1}\Rightarrow t=-1$
Với $x=-\ln \left( e+1 \right)\Rightarrow t=1$
$\begin{aligned}
& I=-\int\limits_{-1}^{1}{f\left( t \right)dt}=-\int\limits_{-1}^{1}{f\left( x \right)dx}=-\int\limits_{-1}^{0}{f\left( x \right)dx}-\int\limits_{0}^{1}{f\left( x \right)dx} \\
& =-\int\limits_{-1}^{0}{\left( -{{x}^{3}}+x \right)dx}-\int\limits_{0}^{1}{\left( {{e}^{x}}-1 \right)dx}=\dfrac{1}{4}-\left( e-2 \right)=\dfrac{9}{4}-e\Rightarrow m=\dfrac{9}{4};n=1\Rightarrow P=2m+\dfrac{n}{2}=5. \\
\end{aligned}$