Ta có \(\int\limits_0^1 {f'\left( x \right){\rm{d}}x}  = f\left( x \right)\left| {\begin{array}{*{20}{c}}
1\\
0
\end{array}} \right. = f\left( 1 \right) - f\left( 0 \right).\)

\(2f\left( x \right) + 3f\left( {1 - x} \right) = \sqrt {1 - {x^2}}  \to \left\{ \begin{array}{l}
2f\left( 0 \right) + 3f\left( 1 \right) = 1\\
2f\left( 1 \right) + 3f\left( 0 \right) = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
f\left( 0 \right) =  - \frac{2}{5}\\
f\left( 1 \right) = \frac{3}{5}
\end{array} \right..\)

\(I = \int\limits_0^1 {f'\left( x \right){\rm{d}}x}  = f\left( 1 \right) - f\left( 0 \right) = \frac{3}{5} + \frac{2}{5} = 1.\)

Ta có \(\int\limits_0^1 {{e^x}\left[ {f\left( x \right) + f'\left( x \right)} \right]{\rm{d}}x}  = \int\limits_0^1 {\left[ {{e^x}f\left( x \right)} \right]{\rm{'d}}x}  = \left[ {{e^x}f\left( x \right)} \right]\left| {_0^1} \right. = ef\left( 1 \right) - f\left( 0 \right)\mathop  = \limits^{f\left( 0 \right) = f\left( 1 \right) = 1} e - 1.\)

\(\left\{ \begin{array}{l}
a = 1\\
b =  - 1
\end{array} \right. \to Q = {a^{2018}} + {b^{2018}} = {1^{2018}} + {\left( { - 1} \right)^{2018}} = 2.\)

\(I = \int\limits_0^2 {{{\left[ {f\left( x \right)g\left( x \right)} \right]}^/}{\rm{d}}x = \int\limits_0^2 {\left[ {f'\left( x \right)g\left( x \right) + f\left( x \right)g'\left( x \right)} \right]{\rm{d}}x} } \)

\( = \int\limits_0^2 {f'\left( x \right)g\left( x \right){\rm{d}}x}  + \,\int\limits_0^2 {f\left( x \right)g'\left( x \right){\rm{d}}x}  = 2 + 3 = 5.\)

Từ \(\int\limits_0^{{x^2}} {f\left( t \right){\rm{d}}t}  = x.\sin \left( {\pi x} \right)\), đạo hàm hai vế ta được \(2xf\left( {{x^2}} \right) = \sin \left( {\pi x} \right) + \pi x\cos \left( {\pi x} \right).\)

Cho \(x = \frac{1}{2}\) ta được \(2.\frac{1}{2}.f\left( {\frac{1}{4}} \right) = \sin \frac{\pi }{2} + \frac{\pi }{2}\cos \frac{\pi }{2} = 1 \to f\left( {\frac{1}{4}} \right) = 1.\)

Từ \(\int\limits_a^x {\frac{{f\left( t \right)}}{{{t^2}}}{\rm{d}}t}  + 6 = 2\sqrt x \), đạo hàm hai vế ta được \(\frac{{f\left( x \right)}}{{{x^2}}} = \frac{1}{{\sqrt x }}.\)

Suy ra \(f\left( x \right) = x\sqrt x  \to f\left( 4 \right) = 4\sqrt 4  = 8.\)

Đặt \(t = \ln \left( {{x^2} + 1} \right),\) suy ra \({\rm{d}}t = \frac{{2x{\rm{d}}x}}{{{x^2} + 1}} \to \frac{{x{\rm{d}}x}}{{{x^2} + 1}} = \frac{{{\rm{d}}t}}{2}.\)

Đổi cận: \(\left\{ \begin{array}{l}
x = 0 \to t = 0\\
x = \sqrt {{e^{2017}} - 1}  \to t = 2017
\end{array} \right..\)

Khi đó \(I = \frac{1}{2}\int\limits_0^{2017} {f\left( t \right){\rm{d}}t}  = \frac{1}{2}\int\limits_0^{2017} {f\left( x \right){\rm{d}}x}  = \frac{1}{2}.2 = 1.\)

Xét \(\int\limits_1^9 {\frac{{f\left( {\sqrt x } \right)}}{{\sqrt x }}{\rm{d}}x = 4} .\) Đặt \(t = \sqrt x  \Rightarrow {t^2} = x,\) suy ra \(2t{\rm{d}}t = {\rm{d}}x.\)

Đổi cận \(\left\{ \begin{array}{l}
x = 1 \to t = 1\\
x = 9 \to t = 3
\end{array} \right..\) Suy ra \(4 = \int\limits_1^9 {\frac{{f\left( {\sqrt x } \right)}}{{\sqrt x }}{\rm{d}}x}  = 2\int\limits_1^3 {f\left( t \right)2{\rm{d}}t}  \to \int\limits_1^3 {f\left( t \right){\rm{d}}t}  = 2.\)

Xét \(\int\limits_0^{\frac{\pi }{2}} {f\left( {\sin x} \right)\cos x{\rm{d}}x = 2} .\) Đặt \(u = \sin x,\) suy ra \({\rm{d}}u = \cos x{\rm{d}}x.\)

Đổi cận \(\left\{ \begin{array}{l}
x = 0 \to u = 0\\
x = \frac{\pi }{2} \to u = 1
\end{array} \right..\) Suy ra \(2 = \int\limits_0^{\frac{\pi }{2}} {f\left( {\sin x} \right)\cos x{\rm{d}}x}  = \int\limits_0^1 {f\left( t \right){\rm{d}}t} .\)

Vậy \(I = \int\limits_0^3 {f\left( x \right){\rm{d}}x}  = \int\limits_0^1 {f\left( x \right){\rm{d}}x}  + \int\limits_1^3 {f\left( x \right){\rm{d}}x}  = 4.\)

Xét \(\int\limits_0^{\frac{\pi }{4}} {f\left( {\tan x} \right){\rm{d}}x}  = 4.\)

Đặt \(t=tan x\) suy ra \({\rm{d}}t = \frac{1}{{{{\cos }^2}x}}{\rm{d}}x = \left( {{{\tan }^2}x + 1} \right){\rm{d}}x \to {\rm{d}}x = \frac{{{\rm{d}}t}}{{1 + {t^2}}}.\)

Đổi cận: \(\left\{ \begin{array}{l}
x = 0 \to t = 0\\
x = \frac{\pi }{4} \to t = 1
\end{array} \right..\) Khi đó \(4 = \int\limits_0^{\frac{\pi }{4}} {f\left( {\tan x} \right){\rm{d}}x}  = \int\limits_0^1 {\frac{{f\left( t \right)}}{{{t^2} + 1}}{\rm{d}}} t = \int\limits_0^1 {\frac{{f\left( x \right)}}{{{x^2} + 1}}{\rm{d}}x} .\)

Từ đó suy ra \(I = \int\limits_0^1 {f\left( x \right){\rm{d}}x}  = \int\limits_0^1 {\frac{{f\left( x \right)}}{{{x^2} + 1}}{\rm{d}}x}  + \int\limits_0^1 {\frac{{{x^2}f\left( x \right)}}{{{x^2} + 1}}{\rm{d}}x}  = 4 + 2 = 6.\)

Đặt \(x = \frac{1}{t},\) suy ra \({\rm{d}}x =  - \frac{1}{{{t^2}}}{\rm{d}}t.\) Đổi cận: \(\left\{ \begin{array}{l}
x = \frac{1}{2} \to t = 2\\
x = 2 \to t = \frac{1}{2}
\end{array} \right..\)

Khi đó \(I = \int\limits_2^{\frac{1}{2}} {\frac{{f\left( {\frac{1}{t}} \right)}}{{\frac{1}{{{t^2}}} + 1}}.\left( { - \frac{1}{{{t^2}}}} \right){\rm{d}}t}  = \int\limits_{\frac{1}{2}}^2 {\frac{{f\left( {\frac{1}{t}} \right)}}{{{t^2} + 1}}{\rm{d}}t}  = \int\limits_{\frac{1}{2}}^2 {\frac{{f\left( {\frac{1}{x}} \right)}}{{{x^2} + 1}}{\rm{d}}x} .\)

Suy ra \(2I = \int\limits_{\frac{1}{2}}^2 {\frac{{f\left( x \right)}}{{{x^2} + 1}}{\rm{d}}x}  + \int\limits_{\frac{1}{2}}^2 {\frac{{f\left( {\frac{1}{x}} \right)}}{{{x^2} + 1}}{\rm{d}}x}  = \int\limits_{\frac{1}{2}}^2 {\frac{{f\left( x \right) + f\left( {\frac{1}{x}} \right)}}{{{x^2} + 1}}{\rm{d}}x}  = \int\limits_{\frac{1}{2}}^2 {\frac{{{x^2} + \frac{1}{{{x^2}}} + 2}}{{{x^2} + 1}}{\rm{d}}x} \)

\( = \int\limits_{\frac{1}{2}}^2 {\frac{{{x^2} + 1}}{{{x^2}}}{\rm{d}}x}  = \int\limits_{\frac{1}{2}}^2 {\left( {1 + \frac{1}{{{x^2}}}} \right){\rm{d}}x}  = \left( {x - \frac{1}{x}} \right)\left| {\begin{array}{*{20}{c}}
2\\
{\frac{1}{2}}
\end{array}} \right. = 3 \to I = \frac{3}{2}.\)

Đặt \(t =  - x \to {\rm{d}}x =  - {\rm{d}}t.\) Đổi cận: \(\left\{ \begin{array}{l}
x =  - \frac{{3\pi }}{2} \to t = \frac{{3\pi }}{2}\\
x = \frac{{3\pi }}{2} \to t =  - \frac{{3\pi }}{2}
\end{array} \right..\)

Khi đó \(I =  - \int\limits_{\frac{{3\pi }}{2}}^{ - \frac{{3\pi }}{2}} {f\left( { - t} \right){\rm{d}}t}  = \int\limits_{ - \frac{{3\pi }}{2}}^{\frac{{3\pi }}{2}} {f\left( { - t} \right){\rm{d}}t}  = \int\limits_{ - \frac{{3\pi }}{2}}^{\frac{{3\pi }}{2}} {f\left( { - x} \right){\rm{d}}x} .\)

Suy ra \(2I = \int\limits_{ - \frac{{3\pi }}{2}}^{\frac{{3\pi }}{2}} {\left[ {f\left( t \right) + f\left( { - t} \right)} \right]{\rm{d}}t}  = \int\limits_{ - \frac{{3\pi }}{2}}^{\frac{{3\pi }}{2}} {\sqrt {2 + 2\cos 2t} {\rm{d}}t}  = \int\limits_{ - \frac{{3\pi }}{2}}^{\frac{{3\pi }}{2}} {2\left| {\cos t} \right|{\rm{d}}t} \mathop  = \limits^{{\rm{CASIO}}} 12 \to I = 6.\)

Đặt \(x = {t^5} + 4t + 3,\) suy ra \({\rm{d}}x = \left( {5{t^4} + 4} \right){\rm{d}}t.\) Đổi cận \(\left\{ \begin{array}{l}
x =  - 2 \to t =  - 1\\
x = 8 \to t = 1
\end{array} \right..\)

Khi đó \(\int\limits_{ - 2}^8 {f\left( x \right){\rm{d}}x}  = \int\limits_{ - 1}^1 {f\left( {{t^5} + 4t + 3} \right)\left( {5{t^4} + 4} \right){\rm{d}}t}  = \int\limits_{ - 1}^1 {\left( {2t + 1} \right)\left( {5{t^4} + 4} \right){\rm{d}}t}  = 10.\)

Đặt \(u = f\left( x \right)\), ta thu được \({u^3} + u = x.\) Suy ra \(\left( {3{u^2} + 1} \right){\rm{d}}u = {\rm{d}}x.\)

Từ \({u^3} + u = x\), ta đổi cận \(\left\{ \begin{array}{l}
x = 0 \to u = 0\\
x = 2 \to u = 1
\end{array} \right..\) Khi đó \(I = \int\limits_0^1 {u\left( {3{u^2} + 1} \right){\rm{d}}u}  = \frac{5}{4}.\)

Đặt \(\left\{ \begin{array}{l}
u = x\\
{\rm{d}}v = f'\left( x \right).{e^{f\left( x \right)}}{\rm{d}}x
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{\rm{d}}u = {\rm{d}}x\\
v = {e^{f\left( x \right)}}
\end{array} \right..\) Khi đó \(\int\limits_0^3 {x.f'\left( x \right).{e^{f\left( x \right)}}{\rm{d}}x}  = x.{e^{f\left( x \right)}}\left| {\begin{array}{*{20}{c}}
3\\
0
\end{array}} \right. - \int\limits_0^3 {{e^{f\left( x \right)}}{\rm{d}}x} .\)

Suy ra \(8 = 3.{e^{f\left( 3 \right)}} - \int\limits_0^3 {{e^{f\left( x \right)}}{\rm{d}}x}  \to \int\limits_0^3 {{e^{f\left( x \right)}}{\rm{d}}x}  = 9 - 8 = 1.\)

Xét \(\int\limits_0^{\frac{\pi }{2}} {f'\left( x \right){{\cos }^2}x{\rm{d}}x}  = 10\), đặt \(\left\{ \begin{array}{l}
u = {\cos ^2}x\\
{\rm{d}}v = f'\left( x \right){\cos ^2}x{\rm{d}}x
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{\rm{d}}u =  - \sin 2x{\rm{d}}x\\
v = f\left( x \right)
\end{array} \right..\)

Khi đó \(10 = \int\limits_0^{\frac{\pi }{2}} {f'\left( x \right){{\cos }^2}x{\rm{d}}x}  = {\cos ^2}xf\left( x \right)\left| {\begin{array}{*{20}{c}}
{\frac{\pi }{2}}\\
0
\end{array}} \right. + \int\limits_0^{\frac{\pi }{2}} {f\left( x \right)\sin 2x{\rm{d}}x} {\rm{ }}\)

\( \Leftrightarrow 10 =  - f\left( 0 \right) + \int\limits_0^{\frac{\pi }{2}} {f\left( x \right)\sin 2x{\rm{d}}x}  \to \int\limits_0^{\frac{\pi }{2}} {f\left( x \right)\sin 2x{\rm{d}}x}  = 10 + f\left( 0 \right) = 13.\)

Ta có \(\int\limits_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}} {\left( {2 - \sin 2x} \right){e^{2\cot x}}{\rm{d}}x}  = 2\int\limits_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}} {{e^{2\cot x}}{\rm{d}}x}  - \int\limits_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}} {\sin 2x{e^{2\cot x}}{\rm{d}}x} .\)         (1)

Xét \(\int\limits_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}} {\sin 2x{e^{2\cot x}}{\rm{d}}x}  = \int\limits_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}} {{e^{2\cot x}}{\rm{d}}\left( {{{\sin }^2}x} \right)}  = {\sin ^2}x.{e^{2\cot x}}\left| {_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}}} \right. - \int\limits_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}} {{{\sin }^2}x\left( { - \frac{2}{{{{\sin }^2}x}}} \right){e^{2\cot x}}{\rm{d}}x} \)

\( = {\sin ^2}x.{e^{2\cot x}}\left| {_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}}} \right. + 2\int\limits_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}} {{e^{2\cot x}}{\rm{d}}x} .\)     (2)

Từ (1) và (2) suy ra \(I = {\sin ^2}x.{e^{2\cot x}}\left| {_{\frac{\pi }{{4 + {m^2}}}}^{\frac{\pi }{2}}} \right. =  - 1 + {\sin ^2}\frac{\pi }{{4 + {m^2}}}.{e^{2\cot \frac{\pi }{{4 + {m^2}}}}}.\)

\( \to S = \ln \left( {{{\sin }^2}\frac{\pi }{{4 + {m^2}}}.{e^{2\cot \frac{\pi }{{4 + {m^2}}}}}} \right) = 2cot\left( {\frac{\pi }{{4 + {m^2}}}} \right) + 2\ln \left( {\sin \frac{\pi }{{4 + {m^2}}}} \right).\)

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 \(I = \left( {x + 3} \right)\ln \left( {9 - {x^2}} \right)\left| {\begin{array}{*{20}{c}}
2\\
1
\end{array}} \right. + 2\int\limits_1^2 {\frac{{x\left( {x + 3} \right)}}{{9 - {x^2}}}{\rm{d}}x}  = 5\ln 5 - 4\ln 8 + 2\int\limits_1^2 {\left( { - 1 + \frac{3}{{3 - x}}} \right){\rm{d}}x} \)

\( = 5\ln 5 - 12\ln 2 - 2\left( {x + 3\ln \left| {3 - x} \right|} \right)\left| {\begin{array}{{20}{c}}
2\\
1
\end{array}} \right. = 5\ln 5 - 6\ln 2 - 2 \to \left\{ \begin{array}{l}
a = 5\\
b =  - 6\\
c =  - 2
\end{array} \right. \to P = 13.\)

Ta có \(I = \int\limits_0^{\frac{\pi }{2}} {\frac{{\left( {{x^2} + 2x\cos x + {{\cos }^2}x} \right) + \left( {1 - \sin x} \right)}}{{x + \cos x}}{\rm{d}}x} \)

\(\begin{array}{l}
 = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\left( {x + \cos x} \right)}^2}}}{{x + \cos x}}{\rm{d}}x}  + \int\limits_0^{\frac{\pi }{2}} {\frac{{1 - \sin x}}{{x + \cos x}}{\rm{d}}x}  = \int\limits_0^{\frac{\pi }{2}} {\left( {x + \cos x} \right){\rm{d}}x}  + \int\limits_0^{\frac{\pi }{2}} {\frac{{{\rm{d}}\left( {x + \cos x} \right)}}{{x + \cos x}}} \\
 = \left( {\frac{1}{2}{x^2} + \sin x + \ln \left| {x + \cos x} \right|} \right)\left| \begin{array}{l}
^{\frac{\pi }{2}}\\
_0
\end{array} \right. = \frac{1}{8}{\pi ^2} + 1 + \ln \frac{\pi }{2} = \frac{1}{8}{\pi ^2} + 1 - \ln \frac{2}{\pi }\\
 \to \left\{ \begin{array}{l}
a = \frac{1}{8}\\
b = 1\\
c = 2
\end{array} \right. \to P = a{c^3} + b = 2.
\end{array}\)

Ta có \(I = \int\limits_{\ln \sqrt 3 }^{\ln \sqrt 8 } {\frac{1}{{\sqrt {{e^{2x}} + 1}  - {e^x}}}{\rm{d}}x}  = \int\limits_{\ln \sqrt 3 }^{\ln \sqrt 8 } {\left( {\sqrt {{e^{2x}} + 1}  + {e^x}} \right){\rm{d}}x}  = \int\limits_{\ln \sqrt 3 }^{\ln \sqrt 8 } {\sqrt {{e^{2x}} + 1} {\rm{d}}x}  + \int\limits_{\ln \sqrt 3 }^{\ln \sqrt 8 } {{e^x}{\rm{d}}x} .\)

\(\int\limits_{\ln \sqrt 3 }^{\ln \sqrt 8 } {{e^x}{\rm{d}}x}  = {e^x}\left| \begin{array}{l}
^{\ln \sqrt 8 }\\
_{\ln \sqrt 3 }
\end{array} \right. = 2\sqrt 2  - \sqrt 3 .\)

\(\int\limits_{\ln \sqrt 3 }^{\ln \sqrt 8 } {\sqrt {{e^{2x}} + 1} {\rm{d}}x} .\) Đặt \(t = \sqrt {{e^{2x}} + 1}  \Leftrightarrow {t^2} = {e^{2x}} + 1\) suy ra \(2t{\rm{d}}t = 2{e^{2x}}{\rm{d}}x \Leftrightarrow {\rm{d}}x = \frac{{t{\rm{d}}t}}{{{e^{2x}}}} = \frac{{t{\rm{d}}t}}{{{t^2} - 1}}.\)

Đổi cận \(\left\{ \begin{array}{l}
x = \ln \sqrt 3  \to t = 2\\
x = \ln \sqrt 8  \to t = 3
\end{array} \right..\)

Khi đó \(\int\limits_{\ln \sqrt 3 }^{\ln \sqrt 8 } {\sqrt {{e^{2x}} + 1} {\rm{d}}x}  = \int\limits_2^3 {\frac{{{t^2}{\rm{d}}t}}{{{t^2} - 1}}} dt = \int\limits_2^3 {\left( {1 + \frac{1}{{{t^2} - 1}}} \right){\rm{d}}t}  = \left( {t + \frac{1}{2}\ln \left| {\frac{{t - 1}}{{t + 1}}} \right|} \right)\left| \begin{array}{l}
^3\\
_2
\end{array} \right. = 1 + \frac{1}{2}\ln \frac{3}{2}.\)

Vậy \(I = 1 + \frac{1}{2}\ln \frac{3}{2} + 2\sqrt 2  - \sqrt 3  \to \left\{ \begin{array}{l}
a = 2\\
b = 3
\end{array} \right. \to P = a + b = 5.\)

Ta có \(I = \int\limits_1^2 {\frac{{{\rm{d}}x}}{{\sqrt {x\left( {x + 1} \right)} \left( {\sqrt {x + 1}  + \sqrt x } \right)}}}  = \int\limits_1^2 {\frac{{\sqrt {x + 1}  + \sqrt x }}{{\sqrt {x\left( {x + 1} \right)} {{\left( {\sqrt {x + 1}  + \sqrt x } \right)}^2}}}} \,{\rm{d}}x.\)

Đặt \(u = \sqrt {x + 1}  + \sqrt x \), suy ra \({\rm{d}}u = \left( {\frac{1}{{2\sqrt {x + 1} }} + \frac{1}{{2\sqrt x }}} \right)\,{\rm{d}}x \to 2{\rm{d}}u = \frac{{\sqrt x  + \sqrt {x + 1} }}{{\sqrt {x\left( {x + 1} \right)} }}\,{\rm{d}}x.\)

Đổi cận \(\left\{ \begin{array}{l}
x = 2 \to u = \sqrt 3  + \sqrt 2 \\
x = 1 \to u = \sqrt 2  + 1
\end{array} \right..\) Khi đó \(I = 2\int\limits_{\sqrt 2  + 1}^{\sqrt 3  + \sqrt 2 } {\frac{{{\rm{d}}u}}{{{u^2}}}} \, = \left. { - \frac{2}{u}} \right|_{\sqrt 2  + 1}^{\sqrt 3  + \sqrt 2 } =  - 2\left( {\frac{1}{{\sqrt 3  + \sqrt 2 }} - \frac{1}{{\sqrt 2  + 1}}} \right)\)

\( =  - 2\left( {\frac{{\sqrt 3  - \sqrt 2 }}{{3 - 2}} - \frac{{\sqrt 2  - 1}}{{2 - 1}}} \right) = \sqrt {32}  - \sqrt {12}  - 2 \to \left\{ \begin{array}{l}
a = 32\\
b = 12\\
c = 2
\end{array} \right. \to P = 46.\)

Ta có \(I = \int\limits_0^{\frac{\pi }{4}} {\frac{{\sin 4x}}{{\sqrt {{{\cos }^2}x + 1}  + \sqrt {{{\sin }^2}x + 1} }}{\rm{d}}x}  = \sqrt 2 \int\limits_0^{\frac{\pi }{4}} {\frac{{2\sin 2x\cos 2x}}{{\sqrt {3 + \cos 2x}  + \sqrt {3 - \cos 2x} }}{\rm{d}}x} .\)

Đặt \(t = \cos 2x \to {\rm{d}}t =  - 2\sin 2x{\rm{d}}x.\) Đổi cận: \(\left\{ \begin{array}{l}
x = 0 \to t = 1\\
x = \frac{\pi }{4} \to t = 0
\end{array} \right..\)

Khi đó \(I =  - \sqrt 2 \int\limits_1^0 {\frac{t}{{\sqrt {3 + t}  + \sqrt {3 - t} }}{\rm{d}}t}  = \sqrt 2 \int\limits_0^1 {\frac{t}{{\sqrt {3 + t}  + \sqrt {3 - t} }}{\rm{d}}t}  = \frac{1}{{\sqrt 2 }}\int\limits_0^1 {\left( {\sqrt {3 + t}  - \sqrt {3 - t} } \right){\rm{d}}t} \)

\( = \frac{1}{{\sqrt 2 }}\left[ {\frac{2}{3}\sqrt {{{\left( {3 + t} \right)}^3}}  + \frac{2}{3}\sqrt {{{\left( {3 - t} \right)}^3}} } \right]\left| {\begin{array}{*{20}{c}}
1\\
0
\end{array}} \right. = \frac{{16\sqrt 2  - 12\sqrt 6  + 8}}{6} \to \left\{ \begin{array}{l}
a = 16\\
b =  - 12\\
c = 8
\end{array} \right. \to P = 36.\)

Ta có \(\int\limits_1^4 {\sqrt {\frac{1}{{4x}} + \frac{{\sqrt x  + {e^x}}}{{\sqrt x {e^{2x}}}}} {\rm{d}}x}  = \int\limits_1^4 {\sqrt {\frac{{{e^{2x}} + 4x + 4{e^x}\sqrt x }}{{4x{e^{2x}}}}} {\rm{d}}x}  = \int\limits_1^4 {\sqrt {\frac{{{{\left( {{e^x} + 2\sqrt x } \right)}^2}}}{{{{\left( {2{e^x}\sqrt x } \right)}^2}}}} } {\rm{d}}x\)

\( = \int\limits_1^4 {\frac{{{e^x} + 2\sqrt x }}{{2{e^x}\sqrt x }}{\rm{d}}x}  = \int\limits_1^4 {\left( {\frac{1}{{2\sqrt x }} + \frac{1}{{{e^x}}}} \right){\rm{d}}x}  = \left( {\sqrt x  - \frac{1}{{{e^x}}}} \right)\left| \begin{array}{l}
^4\\
_1
\end{array} \right. = 1 - \frac{1}{{{e^4}}} + \frac{1}{e} = 1 + {e^{ - 1}} - {e^{ - 4}}\)

\( \to \left\{ \begin{array}{l}
a = 1\\
b =  - 1\\
c =  - 4
\end{array} \right. \to P = a + b + c =  - 4.\)

Ta có \(\int\limits_1^e {\frac{{{{\ln }^2}x + \ln x}}{{{{\left( {\ln x + x + 1} \right)}^3}}}{\rm{d}}x}  = \int\limits_1^e {\frac{{\ln x + 1}}{{\ln x + x + 1}}.\frac{{\ln x}}{{{{\left( {\ln x + x + 1} \right)}^2}}}{\rm{d}}x} .\)

Đặt \(t = \frac{{\ln x + 1}}{{\ln x + x + 1}} \to {\rm{d}}t = \left( {\frac{{\ln x + 1}}{{\ln x + x + 1}}} \right){\rm{'d}}x =  - \frac{{\ln x}}{{{{\left( {\ln x + x + 1} \right)}^2}}}{\rm{d}}x.\)

Đổi cận: \(\left\{ \begin{array}{l}
x = 1 \to t = \frac{1}{2}\\
x = e \to t = \frac{2}{{e + 2}}
\end{array} \right..\) Khi đó \(I =  - \int\limits_{\frac{1}{2}}^{\frac{2}{{e + 2}}} {t{\rm{d}}t}  =  - \frac{1}{2}{t^2}\left| \begin{array}{l}
^{\frac{2}{{e + 2}}}\\
_{\frac{1}{2}}
\end{array} \right. = \frac{1}{8} - \frac{2}{{{{\left( {e + 2} \right)}^2}}}.\)

Ta có \(I = \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{x\cos x}}{{\sqrt {1 + {x^2}}  + x}}{\rm{d}}x}  = \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {x\cos x\left( {\sqrt {1 + {x^2}}  - x} \right){\rm{d}}x}  = \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {x\left( {\sqrt {1 + {x^2}}  - x} \right)\cos x{\rm{d}}x} .\)

Lại có \(I = \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{x\cos x}}{{\sqrt {1 + {x^2}}  + x}}{\rm{d}}x} \mathop  = \limits^{x =  - t} \int\limits_{\frac{\pi }{6}}^{ - \frac{\pi }{6}} {\frac{{\left( { - t} \right)\cos \left( { - t} \right)}}{{\sqrt {1 + {{\left( { - t} \right)}^2}}  - t}}{\rm{d}}\left( { - t} \right) = \int\limits_{\frac{\pi }{6}}^{ - \frac{\pi }{6}} {\frac{{t\cos t}}{{\sqrt {1 + {t^2}}  - t}}{\rm{d}}t} } \)

\( =  - \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {t\left( {\sqrt {1 + {t^2}}  + t} \right)\cos t{\rm{d}}t}  =  - \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {x\left( {\sqrt {1 + {x^2}}  + x} \right)\cos x{\rm{d}}x} .\)

Suy ra \(2I = \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {x\left( {\sqrt {1 + {x^2}}  - x} \right)\cos x{\rm{d}}x}  - \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {x\left( {\sqrt {1 + {x^2}}  + x} \right)\cos x{\rm{d}}x}  =  - 2\int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {{x^2}\cos x{\rm{d}}x} \)

\( \to I =  - \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {{x^2}\cos x{\rm{d}}x} .\) Tích phân từng phần hai lần ta được \(I = 2 + \frac{{{\pi ^2}}}{{ - 36}} + \frac{{\sqrt 3 \pi }}{{ - 3}}\)

\( \to \left\{ \begin{array}{l}
a = 2\\
b =  - 36\\
c =  - 3
\end{array} \right. \to P = a - b + c = 35.\)

\(I = \int\limits_{ - 1}^0 {f\left( x \right){\rm{d}}x}  + \int\limits_0^2 {f\left( x \right){\rm{d}}x}  = \int\limits_{ - 1}^0 {{e^{2x}}{\rm{d}}x}  + \int\limits_0^2 {\left( {x + 1} \right){\rm{d}}x}  = \frac{{9{e^2} - 1}}{{2{e^2}}}.\)

Ta có \(f'\left( x \right) = \frac{2}{{2x - 1}}\)

\( \to f\left( x \right) = \int {\frac{2}{{2x - 1}}{\rm{d}}x}  = \ln \left| {2x - 1} \right| + C = \left\{ {\begin{array}{*{20}{l}}
{\ln \left( {1 - 2x} \right) + {C_1}}&{;x < \frac{1}{2}}\\
{\ln \left( {2x - 1} \right) + {C_2}}&{;x > \frac{1}{2}}
\end{array}} \right..\)

\(f\left( 0 \right) = 1 \to \ln \left( {1 - 2.0} \right) + {C_1} = 1 \to {C_1} = 1.\)

\(f\left( 1 \right) = 2 \to \ln \left( {2.1 - 1} \right) + {C_2} = 2 \to {C_2} = 2.\)

Do đó \(f\left( x \right) = \left\{ \begin{array}{l}
\ln \left( {1 - 2x} \right) + 1\,\,{\rm{khi}}\,\,x < \frac{1}{2}\\
\ln \left( {2x - 1} \right) + 2\,\,{\rm{khi}}\,\,x > \frac{1}{2}
\end{array} \right. \to \left\{ \begin{array}{l}
f\left( { - 1} \right) = \ln 3 + 1\\
f\left( 3 \right) = \ln 5 + 2
\end{array} \right.\)

\( \to f\left( { - 1} \right) + f\left( 3 \right) = 3 + \ln 5 + \ln 3 = 3 + \ln 15.\)

Với \(x\) thuộc vào mỗi khoảng \(\left( { - \frac{\pi }{4} + k\pi ; - \frac{\pi }{4} + k\pi } \right),{\rm{ }}k \in Z\) ta có

\(F\left( x \right) = \int {\frac{{{\rm{d}}x}}{{1 + \sin 2x}} = \int {\frac{{{\rm{d}}x}}{{{{\left( {\sin x + \cos x} \right)}^2}}} = \int {\frac{{{\rm{d}}x}}{{2{{\cos }^2}\left( {x + \frac{\pi }{4}} \right)}} = \frac{1}{2}tan\left( {x + \frac{\pi }{4}} \right)} } }  + C.\)

 \(0; - \frac{\pi }{{12}} \in \left( { - \frac{\pi }{4};\frac{\pi }{4}} \right)\) nên \(F\left( 0 \right) - F\left( { - \frac{\pi }{{12}}} \right) = \frac{1}{2}\tan \left( {x - \frac{\pi }{4}} \right)\left| {\begin{array}{*{20}{c}}
0\\
{ - \frac{\pi }{{12}}}
\end{array}} \right. =  - \frac{1}{2} + \frac{{\sqrt 3 }}{2}\mathop  \to \limits^{F(0) = 1} F\left( { - \frac{\pi }{{12}}} \right) = \frac{3}{2} - \frac{{\sqrt 3 }}{2}.\)

\(\pi ;\frac{{11\pi }}{{12}} \in \left( {\frac{\pi }{4};\frac{{5\pi }}{4}} \right)\) nên \(F\left( \pi  \right) - F\left( {\frac{{11\pi }}{{12}}} \right) = \frac{1}{2}\tan \left( {x - \frac{\pi }{4}} \right)\left| {\begin{array}{*{20}{c}}
\pi \\
{\frac{{11\pi }}{{12}}}
\end{array}} \right. =  - \frac{1}{2} + \frac{{\sqrt 3 }}{2}\mathop  \to \limits^{F\left( \pi  \right) = 0} F\left( {\frac{{11\pi }}{{12}}} \right) = \frac{1}{2} - \frac{{\sqrt 3 }}{2}.\)

Vậy \(P = F\left( { - \frac{\pi }{{12}}} \right) - F\left( {\frac{{11\pi }}{{12}}} \right) = 1.\)

Vì \(f(x)\) là hàm số chẵn nên \(\int\limits_1^3 {f\left( { - 2x} \right){\rm{d}}x}  = \int\limits_1^3 {f\left( {2x} \right){\rm{d}}x}  = 3.\)

Xét \(K = \int\limits_1^3 {f\left( {2x} \right){\rm{d}}x}  = 3.\) Đặt \(t = 2x \to {\rm{d}}t = 2{\rm{d}}x.\) Đổi cận: \(\left\{ \begin{array}{l}
x = 1 \to t = 2\\
x = 3 \to t = 6
\end{array} \right..\)

Khi đó \(K = \frac{1}{2}\int\limits_2^6 {f\left( t \right){\rm{d}}t}  = \frac{1}{2}\int\limits_2^6 {f\left( x \right){\rm{d}}x}  \to \int\limits_2^6 {f\left( x \right){\rm{d}}x}  = 2K = 6.\)

Vậy \(I = \int\limits_{ - 1}^6 {f\left( x \right){\rm{d}}x}  = \int\limits_{ - 1}^2 {f\left( x \right){\rm{d}}x}  + \int\limits_2^6 {f\left( x \right){\rm{d}}x}  = 8 + 6 = 14.\)

Đặt \(t = \left( {3 + 7} \right) - x \to {\rm{d}}t =  - {\rm{d}}x.\) Đổi cận \(\left\{ \begin{array}{l}
x = 7 \to t = 3\\
x = 3 \to t = 7
\end{array} \right..\)

Khi đó \(I =  - \int\limits_7^3 {\left( {10 - t} \right)f\left( {10 - t} \right){\rm{d}}t}  = \int\limits_3^7 {\left( {10 - t} \right)f\left( {10 - t} \right){\rm{d}}t}  = \int\limits_3^7 {\left( {10 - x} \right)f\left( {10 - x} \right){\rm{d}}x} \)

\(\mathop  = \limits^{f\left( x \right) = f\left( {10 - x} \right)} \int\limits_3^7 {\left( {10 - x} \right)f\left( x \right){\rm{d}}x}  = 10\int\limits_3^7 {f\left( x \right){\rm{d}}x}  - \int\limits_3^7 {xf\left( x \right){\rm{d}}x}  = 10\int\limits_3^7 {f\left( x \right){\rm{d}}x}  - I.\)

Suy ra \(2I = 10\int\limits_3^7 {f\left( x \right){\rm{d}}x}  = 10.4 = 40 \to I = 20.\)

Đặt \(x =  - t \to {\rm{d}}x =  - {\rm{d}}t.\) Đổi cận \(\left\{ \begin{array}{l}
x =  - \pi  \to t = \pi \\
x = \pi  \to t =  - \pi 
\end{array} \right..\)

Khi đó \(I =  - \int\limits_\pi ^{ - \pi } {\frac{{f\left( { - t} \right)}}{{{{2018}^{ - t}} + 1}}{\rm{d}}t}  = \int\limits_{ - \pi }^\pi  {\frac{{f\left( { - t} \right)}}{{{{2018}^{ - t}} + 1}}{\rm{d}}t}  = \int\limits_{ - \pi }^\pi  {\frac{{{{2018}^t}f\left( { - t} \right)}}{{1 + {{2018}^t}}}{\rm{d}}t}  = \int\limits_{ - \pi }^\pi  {\frac{{{{2018}^x}f\left( { - x} \right)}}{{1 + {{2018}^x}}}{\rm{d}}x} .\)

Vì \(y = f\left( x \right)\) là hàm số chẵn trên đoạn \(\left[ { - \pi ;\pi } \right]\) nên \(f\left( { - x} \right) = f\left( x \right) \to I = \int\limits_{ - \pi }^\pi  {\frac{{{{2018}^x}f\left( x \right)}}{{{{2018}^x} + 1}}{\rm{d}}x} .\)

Vậy \(2I = \int\limits_{ - \pi }^\pi  {\frac{{f\left( x \right)}}{{{{2018}^x} + 1}}{\rm{d}}x}  + \int\limits_{ - \pi }^\pi  {\frac{{{{2018}^x}f\left( x \right)}}{{{{2018}^x} + 1}}{\rm{d}}x}  = \int\limits_{ - \pi }^\pi  {f\left( x \right){\rm{d}}x}  = 2\int\limits_0^\pi  {f\left( x \right){\rm{d}}x}  = 2.2018 \to I = 2018.\)

Gọi \(I = \int\limits_0^\pi  {\frac{{x{{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x} \)

Đặt \(t = \pi  - x \to {\rm{d}}t =  - {\rm{d}}x.\) Đổi cận \(\left\{ \begin{array}{l}
x = 0 \to t = \pi \\
x = \pi  \to t = 0
\end{array} \right..\)

Khi đó \(I =  - \int\limits_\pi ^0 {\frac{{\left( {\pi  - t} \right){{\sin }^{2018}}\left( {\pi  - t} \right)}}{{{{\sin }^{2018}}\left( {\pi  - t} \right) + {{\cos }^{2018}}\left( {\pi  - t} \right)}}{\rm{d}}t}  = \int\limits_0^\pi  {\frac{{\left( {\pi  - t} \right){{\sin }^{2018}}t}}{{{{\sin }^{2018}}t + {{\cos }^{2018}}t}}{\rm{d}}t}  = \int\limits_0^\pi  {\frac{{\left( {\pi  - x} \right){{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x} .\)

Suy ra \(2I = \int\limits_0^\pi  {\frac{{x{{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x}  + \int\limits_0^\pi  {\frac{{\left( {\pi  - x} \right){{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x}  = \int\limits_0^\pi  {\frac{{\pi {{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x} \)

\( \to I = \frac{\pi }{2}\int\limits_0^\pi  {\frac{{{{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x}  = \frac{\pi }{2}\left[ {\int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x}  + \int\limits_{\frac{\pi }{2}}^\pi  {\frac{{{{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x} } \right].\)

Đặt \(x = u + \frac{\pi }{2}\) ta suy ra \(\int\limits_{\frac{\pi }{2}}^\pi  {\frac{{{{\sin }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x}  = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\cos }^{2018}}u}}{{{{\sin }^{2018}}u + {{\cos }^{2018}}u}}{\rm{d}}u}  = \int\limits_{\frac{\pi }{2}}^\pi  {\frac{{{{\cos }^{2018}}x}}{{{{\sin }^{2018}}x + {{\cos }^{2018}}x}}{\rm{d}}x} .\)

Vậy \(I = \frac{\pi }{2}\int\limits_0^{\frac{\pi }{2}} {{\rm{d}}x}  = \frac{{{\pi ^2}}}{4} \to \left\{ \begin{array}{l}
a = 2\\
b = 4
\end{array} \right. \to P = 8.\)