Tích phân (phương pháp đổi biến)

$\int\limits_{ - a}^a {f\left( x \right)dx}  = \int\limits_{ - a}^0 {f\left( x \right)dx}  + \int\limits_0^a {f\left( x \right)dx}$

Đặt $x =  - t$ thì $dx =  - dt$ $\Rightarrow \int\limits_{ - a}^0 {f\left( x \right)dx}  = \int\limits_a^0 {f\left( { - t} \right)\left( { - dt} \right)}  = \int\limits_0^a {f\left( { - t} \right)dt}$

Mà $f\left( x \right)$ là hàm chẵn nên $f\left( { - t} \right) = f\left( t \right)$ hay $\int\limits_0^a {f\left( { - t} \right)dt}  = \int\limits_0^a {f\left( t \right)dt}  = \int\limits_0^a {f\left( x \right)dx}$

Do đó $\int\limits_{ - a}^0 {f\left( x \right)dx}  = \int\limits_0^a {f\left( x \right)dx}$ $\Rightarrow \int\limits_{ - a}^a {f\left( x \right)dx}  = \int\limits_{ - a}^0 {f\left( x \right)dx}  + \int\limits_0^a {f\left( x \right)dx}  = 2\int\limits_0^a {f\left( x \right)dx}$

Ta có: $I = \int\limits_e^{{e^2}} {\dfrac{{dx}}{{x\ln x\ln ex}}}  = \int\limits_e^{{e^2}} {\dfrac{{dx}}{{x\ln x\left( {1 + \ln x} \right)}}}$

Đặt $t = \ln x \Leftrightarrow dt = \dfrac{{dx}}{x}$

Đổi cận: $\left\{ \begin{array}{l}x = e \Leftrightarrow t = 1\\x = {e^2} \Leftrightarrow t = 2\end{array} \right.$, khi đó

$\begin{array}{l}I = \int\limits_1^2 {\dfrac{{dt}}{{t\left( {t + 1} \right)}}}  = \int\limits_1^2 {\left( {\dfrac{1}{t} - \dfrac{1}{{t + 1}}} \right)dx}  = \left. {\left( {\ln \left| t \right| - \ln \left| {t + 1} \right|} \right)} \right|_1^2 = \left. {\ln \left| {\dfrac{t}{{t + 1}}} \right|} \right|_1^2\\\,\,\, = \ln \dfrac{2}{3} - \ln \dfrac{1}{2} = \ln \dfrac{4}{3} = \ln \dfrac{a}{b} \Leftrightarrow \left\{ \begin{array}{l}a = 4\\b = 3\end{array} \right. \Leftrightarrow a - b = 1\end{array}$

Ta có: $\int\limits_{0}^{1}{f\left( x \right)dx=\frac{1}{2}\int\limits_{1}^{2}{f\left( x \right)dx=1\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx=1}}}$ và $\int\limits_{1}^{2}{f\left( x \right)dx=2.}$

$\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx+\int\limits_{1}^{2}{f\left( x \right)dx}=\int\limits_{0}^{2}{f\left( x \right)dx=3.}}$

Mặt khác: $\int\limits_{-2}^{2}{\frac{f\left( x \right)}{{{3}^{x}}+1}dx}=\int\limits_{-2}^{0}{\frac{f\left( x \right)}{{{3}^{x}}+1}dx+\int\limits_{0}^{2}{\frac{f\left( x \right)}{{{3}^{x}}+1}dx}}$ và $y=f\left( x \right)$ là hàm số chẵn, liên tục trên $R.$

$\Rightarrow f\left( -x \right)=f\left( x \right)\ \forall x\in R.$

Gọi $I=\int\limits_{-2}^{2}{\frac{f\left( x \right)}{{{3}^{x}}+1}\,\text{d}x}$, đặt $t=-\,x\Rightarrow \text{d}t=-\,\text{d}x$ và đổi cận $\left\{ \begin{align}  & x=-\,2\,\,\Rightarrow \,\,t=2 \\  & x=2\,\,\Rightarrow \,\,t=-\,2 \\ \end{align} \right..$

Suy ra $I=\int\limits_{2}^{-\,2}{\frac{f\left( -t \right)}{{{3}^{-t}}+1}\,\left( -\,\text{d}t \right)}=\int\limits_{-\,2}^{2}{\frac{f\left( t \right)}{\frac{1}{{{3}^{t}}}+1}\,\text{d}t}=\int\limits_{-\,2}^{2}{\frac{{{3}^{x}}f\left( x \right)}{{{3}^{x}}+1}\,\text{d}x}$

$\Rightarrow \,\,2I=\int\limits_{-\,2}^{2}{\frac{\left( {{3}^{x}}+1 \right)f\left( x \right)}{{{3}^{x}}+1}\,\text{d}x}=\int\limits_{-\,2}^{2}{f\left( x \right)\,\text{d}x}$

Do $f\left( x \right)$ là hàm chẵn nên suy ra $\int\limits_{-\,2}^{2}{f\left( x \right)\,\text{d}x}=2\int\limits_{0}^{2}{f\left( x \right)\,\text{d}x}$.

Vậy $I=\int\limits_{0}^{2}{f\left( x \right)\,\text{d}x}=\int\limits_{0}^{1}{f\left( x \right)\,\text{d}x}+\int\limits_{1}^{2}{f\left( x \right)\,\text{d}x}=3.$

Đặt $t = \ln x \Leftrightarrow dt = \dfrac{{dx}}{x}$

Đổi cận $\left\{ \begin{array}{l}x = 1 \Leftrightarrow t = 0\\x = e \Leftrightarrow t = 1\end{array} \right.$, khi đó ta có

$\begin{array}{l}I = \int\limits_0^1 {\dfrac{{tdt}}{{{{\left( {2 + t} \right)}^2}}}}  = \int\limits_0^1 {\dfrac{{2 + t - 2}}{{{{\left( {2 + t} \right)}^2}}}dt}  = \int\limits_0^1 {\dfrac{1}{{2 + t}}dt}  - 2\int\limits_0^1 {\dfrac{1}{{{{\left( {2 + t} \right)}^2}}}dt}  = \\ = \left. {\left( {\ln \left| {2 + t} \right| + \dfrac{2}{{2 + t}}} \right)} \right|_0^1 = \ln 3 + \dfrac{2}{3} - \ln 2 - 1 =  - \dfrac{1}{3} + \ln \dfrac{3}{2} \Leftrightarrow \left\{ \begin{array}{l}a = 3\\b = 2\end{array} \right. \Leftrightarrow 2a - b = 4\end{array}$

Đặt $t = \sqrt {{e^x} + 3}  \Rightarrow {t^2} = {e^x} + 3 \Leftrightarrow 2tdt = {e^x}dx$

Đổi cận: $\left\{ \begin{array}{l}x = 0 \Rightarrow t = 2\\x = b \Rightarrow t = \sqrt {{e^b} + 3} \end{array} \right.$

Khi đó ta có:

$\int\limits_0^b {\frac{{{e^x}}}{{\sqrt {{e^x} + 3} }}dx}  = 2 \Leftrightarrow \int\limits_2^{\sqrt {{e^b} + 3} } {\frac{{2tdt}}{t}}  = 2 \Leftrightarrow \left. t \right|_2^{\sqrt {{e^b} + 3} } = 1 \Leftrightarrow \sqrt {{e^b} + 3}  - 2 = 1 \Leftrightarrow b = \ln 6 \approx 1,8$

Vậy trong các khoảng ở đáp án chỉ có đáp án A thỏa mãn.

Đặt $t = \sqrt {3\tan x + 1}  \Leftrightarrow {t^2} = 3\tan x + 1 \Leftrightarrow 2tdt = \dfrac{3}{{{{\cos }^2}x}}dx$ và $\tan x = \dfrac{{{t^2} - 1}}{3}$

Đổi cận $\left\{ \begin{array}{l}x = 0 \Leftrightarrow t = 1\\x = \dfrac{\pi }{4} \Leftrightarrow t = 2\end{array} \right.$ . Khi đó ta có:

$I = \int\limits_1^2 {\dfrac{{2\tan x.3}}{{{{\cos }^2}x\sqrt {3\tan x + 1} }}dx}  = 2\int\limits_1^2 {\dfrac{{\dfrac{{{t^2} - 1}}{3}.2tdt}}{t}}  = \dfrac{4}{3}\int\limits_1^2 {\left( {{t^2} - 1} \right)dt}$

Ta có $I=\int\limits_{1}^{2}{\frac{x+1}{{{x}^{2}}+x\ln x}\,\text{d}x}=\int\limits_{1}^{2}{\frac{1+\frac{1}{x}}{x+\ln x}\,\text{d}x}=\int\limits_{1}^{2}{\frac{\text{d}\left( x+\ln x \right)}{x+\ln x}}=\left. \ln \left| x+\ln x \right| \right|_{1}^{2}=\ln \left( \ln 2+2 \right).$

Mặt khác $\int\limits_{1}^{2}{\frac{x+1}{{{x}^{2}}+x\ln x}\,\text{d}x}=\ln \left( \ln a+b \right)=\ln \left( \ln 2+2 \right)\Rightarrow \,\,\left\{ \begin{align}  a=2 \\  b=2 \\ \end{align} \right.\Rightarrow \,\,P=12.$

Đặt $\cos x = \tan a \Leftrightarrow  - \sin xdx = \left( {1 + {{\tan }^2}a} \right)da$

Đổi cận $\left\{ \begin{array}{l}x = 0 \Leftrightarrow a = \dfrac{\pi }{4}\\x = \dfrac{\pi }{4} \Leftrightarrow a = \arctan \dfrac{{\sqrt 2 }}{2}\end{array} \right.$, khi đó ta có:   $I = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\tan xdx}}{{1 + {{\cos }^2}x}}}  = \int\limits_{\dfrac{\pi }{4}}^{\arctan \dfrac{{\sqrt 2 }}{2}} {\dfrac{{ - \left( {1 + {{\tan }^2}a} \right)da}}{{\tan a\left( {1 + {{\tan }^2}a} \right)}}}  = \int\limits_{\arctan \dfrac{{\sqrt 2 }}{2}}^{\dfrac{\pi }{4}} {\dfrac{{\cos ada}}{{\sin a}}}$

Đặt $u = \sin a \Leftrightarrow du = \cos ada$, đổi cận $\left\{ \begin{array}{l}a = \dfrac{\pi }{4} \Leftrightarrow u = \dfrac{{\sqrt 2 }}{2}\\a = \arctan \dfrac{{\sqrt 2 }}{2} \Leftrightarrow u = \dfrac{{\sqrt 3 }}{3}\end{array} \right.$ , khi đó ta có:

$\begin{array}{l}I = \int\limits_{\dfrac{{\sqrt 3 }}{3}}^{\dfrac{{\sqrt 2 }}{2}} {\dfrac{{du}}{u}}  = \left. {\ln u} \right|_{\dfrac{{\sqrt 3 }}{3}}^{\dfrac{{\sqrt 2 }}{2}} = \ln \dfrac{{\sqrt 2 }}{2} - \ln \dfrac{{\sqrt 3 }}{3} = \ln \dfrac{{\sqrt 6 }}{2} = \dfrac{1}{2}\ln {\left( {\dfrac{{\sqrt 6 }}{2}} \right)^2} = \dfrac{1}{2}\ln \dfrac{3}{2}\\ \Rightarrow m = \dfrac{1}{2}\end{array}$

Đặt $t=\sqrt[3]{1+{{x}^{2}}}\Leftrightarrow {{t}^{3}}=1+{{x}^{2}}\Leftrightarrow 2x\,\text{d}x=3{{t}^{2}}\,\text{d}t\Leftrightarrow x\,\text{d}x=\frac{3{{t}^{2}}}{2}\,\text{d}t$ và $\left\{\begin{align}  x=0\,\,\Rightarrow \,\,t=1 \\  x=\sqrt{7}\,\,\Rightarrow \,\,t=2 \\ \end{align} \right..$

Khi đó $\int\limits_{0}^{\sqrt{7}}{\frac{{{x}^{3}}\,\text{d}x}{\sqrt[3]{1+{{x}^{2}}}}}=\int\limits_{0}^{\sqrt{7}}{\frac{{{x}^{2}}}{\sqrt[3]{1+{{x}^{2}}}}.x\,\text{d}x}=\int\limits_{1}^{2}{\frac{{{t}^{3}}-1}{t}.\frac{3{{t}^{2}}}{2}\,\text{d}t}=\frac{3}{2}\,\int\limits_{1}^{2}{\left( {{t}^{4}}-t \right)\,\text{d}t}=\frac{141}{20}.$

Vậy $\int\limits_{0}^{\sqrt{7}}{\frac{{{x}^{3}}\,\text{d}x}{\sqrt[3]{1+{{x}^{2}}}}}=\frac{m}{n}\Rightarrow\left\{ \begin{align}  m=141 \\  n=20 \\ \end{align} \right.\,\,\xrightarrow{{}}\,\,m-7n=141-7.20=1.$

Đặt $x = a\tan t \Leftrightarrow dx = a\dfrac{1}{{{{\cos }^2}x}}dx = a\left( {1 + {{\tan }^2}t} \right)dt$

Đổi cận: $\left\{ \begin{array}{l}x = 0 \Leftrightarrow t = 0\\x = a \Leftrightarrow t = \dfrac{\pi }{4}\end{array} \right.$ , khi đó $\int\limits_0^a {\dfrac{1}{{{x^2} + {a^2}}}dx}  = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{a\left( {1 + {{\tan }^2}t} \right)dt}}{{{a^2}{{\tan }^2}t + {a^2}}}}  = \dfrac{1}{a}\int\limits_0^{\dfrac{\pi }{4}} {dt}  = \dfrac{\pi }{{4a}}$

Đặt $u=\sqrt{4x+5}\Leftrightarrow {{u}^{2}}=4x+5\Leftrightarrow 2u\,\text{d}u=4\,\text{d}x\Leftrightarrow \text{d}x=\frac{u}{2}\,\text{d}u.$

Có ${{u}^{2}}=4x+5\Rightarrow x=\frac{{{u}^{2}}-5}{4}.$

Đổi cận : $\left\{ \begin{align}  & x=-\,1\Rightarrow u=1 \\  & x=1\Rightarrow u=3 \\ \end{align} \right..$

Khi đó $\int\limits_{-\,1}^{1}{x\sqrt{4x+5}\,\text{d}x}=\int\limits_{1}^{3}{\frac{{{u}^{2}}-5}{4}.u.\frac{u}{2}\,\text{d}u}=\int\limits_{1}^{3}{\frac{{{u}^{2}}\left( {{u}^{2}}-5 \right)}{8}\,\text{d}u}.$

Đặt $t = \sqrt {{x^2} - 3}  \Leftrightarrow {t^2} = {x^2} - 3 \Leftrightarrow tdt = xdx$ và ${x^2} = {t^2} + 3$

Đổi cận : $\left\{ \begin{array}{l}x = 2 \Leftrightarrow t = 1\\x = 2\sqrt 3  \Leftrightarrow t = 3\end{array} \right.$, khi đó ta có :

$I = \int\limits_2^{2\sqrt 3 } {\dfrac{{\sqrt 3 xdx}}{{{x^2}\sqrt {{x^2} - 3} }}}  = \int\limits_1^3 {\dfrac{{\sqrt 3 tdt}}{{\left( {{t^2} + 3} \right)t}}}  = \sqrt 3 \int\limits_1^3 {\dfrac{{dt}}{{{t^2} + 3}}}$

Đặt $t = \sqrt 3 \tan \alpha  \Leftrightarrow dt = \dfrac{{\sqrt 3 }}{{{{\cos }^2}\alpha }}d\alpha  = \sqrt 3 \left( {1 + {{\tan }^2}\alpha } \right)d\alpha$

Đổi cận : $\left\{ \begin{array}{l}t = 1 \Leftrightarrow \dfrac{\pi }{6}\\t = 3 \Leftrightarrow t = \dfrac{\pi }{3}\end{array} \right.$ , khi đó ta có : $I = \sqrt 3 \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {\dfrac{{\sqrt 3 \left( {1 + {{\tan }^2}\alpha } \right)d\alpha }}{{3{{\tan }^2}\alpha  + 3}}}  = \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} {d\alpha }  = \left. \alpha  \right|_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}} = \dfrac{\pi }{6}$

Đặt $t={{e}^{\sqrt{x\,+\,1}}}\Leftrightarrow2\,\text{d}t=\frac{{{e}^{\sqrt{x\,+\,1}}}}{\sqrt{x+1}}\text{d}x$ và đổi cận $\left\{ \begin{align}  x=0\,\,\Rightarrow \,\,t=e \\  x=3\,\,\Rightarrow \,\,t={{e}^{2}} \\\end{align} \right..$

Khi đó $\int\limits_{0}^{3}{{{e}^{\sqrt{x\,+\,1}}}.\frac{\text{d}x}{\sqrt{x+1}}}=2\,\int\limits_{e}^{{{e}^{2}}}{\text{d}t}=\left. 2t \right|_{e}^{{{e}^{2}}}=2{{e}^{2}}-2e=a.{{e}^{2}}+b.e+c\Rightarrow \left\{ \begin{align}  a=2 \\  b=-\,2 \\  c=0 \\ \end{align} \right..$

Vậy $S=0.$

Đặt $t = \sqrt {{e^x} + 1}  \Leftrightarrow {t^2} = {e^x} + 1 \Leftrightarrow 2tdt = {e^x}dx = \left( {{t^2} - 1} \right)dx \Leftrightarrow dx = \dfrac{{2t}}{{{t^2} - 1}}dt$

Đổi cận: $\left\{ \begin{array}{l}x = \ln 3 \Leftrightarrow t = 2\\x = \ln 8 \Leftrightarrow t = 3\end{array} \right.$, khi đó ta có:

$\begin{array}{l}I = \int\limits_2^3 {\dfrac{{{t^2} - 1 - 1}}{t}\dfrac{{2t}}{{{t^2} - 1}}dt}  = 2\int\limits_2^3 {\dfrac{{{t^2} - 2}}{{{t^2} - 1}}dt}  = 2\int\limits_2^3 {\left( {1 - \dfrac{1}{{\left( {t - 1} \right)\left( {t + 1} \right)}}} \right)dt} \\ = 2\int\limits_2^3 {\left( {1 + \dfrac{1}{2}\dfrac{{t - 1 - \left( {t + 1} \right)}}{{\left( {t - 1} \right)\left( {t + 1} \right)}}} \right)dt}  = 2\int\limits_2^3 {\left( {1 + \dfrac{1}{2}\left( {\dfrac{1}{{t + 1}} - \dfrac{1}{{t - 1}}} \right)} \right)dt} \\ = \left. {\left( {2x + \ln \left( {t + 1} \right) - \ln \left( {t - 1} \right)} \right)} \right|_2^3\\ = 6 + \ln 4 - \ln 2 - 4 - \ln 3 = 2 - \ln \dfrac{3}{2} \Leftrightarrow \left\{ \begin{array}{l}a = 2\\b = 3\end{array} \right. \Leftrightarrow a + b = 5\end{array}$

Đổi cận: $\left\{ \begin{align} & x=1\Rightarrow u=1 \\ & x=e\Rightarrow u=2 \\ \end{align} \right..$

Ta có: $u=\sqrt{1+3\ln x}\Rightarrow {{u}^{2}}=1+3\ln x\Rightarrow \ln x=\frac{{{u}^{2}}-1}{3}.$

$\begin{align} & u=\sqrt{1+3\ln x}\Rightarrow du=\left( \sqrt{1+3\ln x} \right)'dx=\frac{\left( 1+3\ln x \right)'}{2\sqrt{1+3\ln x}}dx=\frac{3}{2x\sqrt{1+3\ln x}}dx. \\ & \Rightarrow \frac{1}{x\sqrt{1+3\ln x}}dx=\frac{2}{3}du \\ \end{align}$ $\Rightarrow \int\limits_{1}^{e}{\frac{\ln x}{x\sqrt{1+3\ln x}}}dx=\int\limits_{1}^{2}{\frac{{{u}^{2}}-1}{3}.\frac{2}{3}du=\frac{2}{9}\int\limits_{1}^{2}{\left( {{u}^{2}}-1 \right)du.}}$

Đặt $t = \sin x + \cos x \Rightarrow dt = \left( {\cos x - \sin x} \right)dx,$ đổi cận $\left\{ \begin{array}{l}x = \dfrac{\pi }{6} \Rightarrow t = \dfrac{{1 + \sqrt 3 }}{2}\\x = \dfrac{\pi }{4} \Rightarrow t = \sqrt 2 \end{array} \right.$

$\Rightarrow I = \int\limits_{\dfrac{{1 + \sqrt 3 }}{2}}^{\sqrt 2 } {\dfrac{{ - dt}}{t}}  = \left. { - \ln \left| t \right|} \right|_{\dfrac{{1 + \sqrt 3 }}{2}}^{\sqrt 2 } =  - \ln \sqrt 2  + \ln \dfrac{{1 + \sqrt 3 }}{2} = \ln \dfrac{{1 + \sqrt 3 }}{{2\sqrt 2 }} = \ln \dfrac{{\sqrt 2  + \sqrt 6 }}{4}$

Đặt $t=\sin 2x\Rightarrow dt=2\cos 2xdx$, đổi cận $\left\{ \begin{align}  x=0\Rightarrow t=0 \\  x=\frac{\pi }{4}\Rightarrow t=1 \\ \end{align} \right.\Rightarrow \int\limits_{0}^{\frac{\pi }{4}}{f\left( \sin 2x \right)\cos 2xdx}=\frac{1}{2}\int\limits_{0}^{1}{f\left( t \right)dt}=\frac{1}{2}.2018=1009$

Đặt $t = \dfrac{{\sqrt {1 + {x^2}} }}{x}$ $\Leftrightarrow {t^2}{x^2} = 1 + {x^2} \Leftrightarrow {x^2}\left( {{t^2} - 1} \right) = 1 \Rightarrow {x^2} = \dfrac{1}{{{t^2} - 1}}$ $\Rightarrow 2xdx = \dfrac{{ - 2t}}{{{{\left( {{t^2} - 1} \right)}^2}}}dt$

$\Rightarrow \dfrac{{dx}}{x} = \dfrac{{ - tdt}}{{{{\left( {{t^2} - 1} \right)}^2}}}.\left( {{t^2} - 1} \right) = \dfrac{{ - tdt}}{{{t^2} - 1}}$

Đổi cận $\left\{ \begin{array}{l}x = 1 \Rightarrow t = \sqrt 2 \\x = \sqrt 3  \Rightarrow t = \dfrac{2}{{\sqrt 3 }}\end{array} \right.$ , khi đó ta có:

$\begin{array}{l}I =  - \int\limits_{\sqrt 2 }^{\dfrac{2}{{\sqrt 3 }}} {\dfrac{{{t^2}dt}}{{{t^2} - 1}}}  = \int\limits_{\dfrac{2}{{\sqrt 3 }}}^{\sqrt 2 } {\left( {1 + \dfrac{1}{{{t^2} - 1}}} \right)dt}  \\= \left( {\sqrt 2  - \dfrac{2}{{\sqrt 3 }}} \right) + \int\limits_{\dfrac{2}{{\sqrt 3 }}}^{\sqrt 2 } {\dfrac{1}{{{t^2} - 1}}dt}  \\= \left( {\sqrt 2  - \dfrac{2}{{\sqrt 3 }}} \right) + \left. {\dfrac{1}{2}\ln \left| {\dfrac{{t - 1}}{{t + 1}}} \right|} \right|_{\dfrac{2}{{\sqrt 3 }}}^{\sqrt 2 }\\ = \left( {\sqrt 2  - \dfrac{2}{{\sqrt 3 }}} \right) + \dfrac{1}{2}\left( {\ln \left( {3 - 2\sqrt 2 } \right) - \ln \left( {7 - 4\sqrt 3 } \right)} \right)\\ = \sqrt 2  - \dfrac{2}{{\sqrt 3 }} + \ln \left( {\sqrt 2  - 1} \right) + \ln \left( {2 - \sqrt 3 } \right)\\ = \sqrt 2  - \dfrac{2}{{\sqrt 3 }} + \ln \dfrac{{\sqrt 2  - 1}}{{2 - \sqrt 3 }}\end{array}$

Đặt $t = 9x \Leftrightarrow dt = 9dx$.

Đổi cận: $\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = 3 \Leftrightarrow t = 27\end{array} \right.$, khi đó ta có:

 $\int\limits_0^3 {f\left( {9x} \right)dx}  = \int\limits_0^{27} {f\left( t \right)\frac{{dt}}{9}}  = \frac{1}{9}\int\limits_0^{27} {f\left( t \right)dt}  = \frac{1}{9}\int\limits_0^{27} {f\left( x \right)dx}  = \frac{1}{9}.81 = 9$

$I = \int\limits_0^{\dfrac{\pi }{6}} {\dfrac{{\sin 2x}}{{{{\left( {2 + \sin x} \right)}^2}}}dx}  = 2\int\limits_0^{\dfrac{\pi }{6}} {\dfrac{{\sin x\cos x}}{{{{\left( {2 + \sin x} \right)}^2}}}dx}$

Đặt $t = \sin x \Rightarrow dt = \cos xdx$, đổi cận $\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = \dfrac{\pi }{6} \Rightarrow t = \dfrac{1}{2}\end{array} \right.$

$\begin{array}{l} \Rightarrow I = 2\int\limits_0^{\dfrac{1}{2}} {\dfrac{{tdt}}{{{{\left( {2 + t} \right)}^2}}}}  = 2\int\limits_0^{\dfrac{1}{2}} {\dfrac{{t + 2 - 2}}{{{{\left( {t + 2} \right)}^2}}}dt = 2\int\limits_0^{\dfrac{1}{2}} {\left( {\dfrac{1}{{t + 2}} - \dfrac{2}{{{{\left( {t + 2} \right)}^2}}}} \right)dt} } \\ = 2\left. {\left( {\ln \left( {t + 2} \right) + \dfrac{2}{{t + 2}}} \right)} \right|_0^{\dfrac{1}{2}} = 2\left( {\ln \dfrac{5}{2} + \dfrac{4}{5} - \ln 2 - 1} \right) = 2\left( {\ln \dfrac{5}{4} - \dfrac{1}{5}} \right) =  - \dfrac{2}{5} + 2\ln \dfrac{5}{4}\\ \Rightarrow \left\{ \begin{array}{l}a =  - 2\\b = 2\end{array} \right. \Rightarrow T = a + b = 0\end{array}$