Tích phân (phương pháp từng phần)

Đặt \(\left\{ \begin{array}{l}
u = 2x - 1\\
{\rm{d}}v = {e^{2x}}\,{\rm{d}}x
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{\rm{d}}u = 2\,{\rm{d}}x\\
v = \frac{1}{2}{e^{2x}}
\end{array} \right.\)

Khi đó ta có: 

\(\begin{array}{l}
I = \frac{1}{2}\left( {2x - 1} \right){e^{2x}}\left| \begin{array}{l}
m\\
0
\end{array} \right. - \int\limits_0^m {{e^{2x}}{\rm{d}}x = \frac{1}{2}\left( {2m - 1} \right){e^{2m}} + \frac{1}{2} - \left. {\frac{1}{2}{e^{2x}}} \right|_0^m} \\
= \frac{1}{2}\left( {2m - 1} \right){e^{2m}} + \frac{1}{2} - \frac{1}{2}{e^{2m}} + \frac{1}{2} = \frac{1}{2}\left( {2m - 2} \right){e^{2m}} + 1.
\end{array}\)

Ta có

\(I<m\Leftrightarrow \frac{1}{2}\left( 2m-2 \right){{e}^{2m}}+1<m\Leftrightarrow \left( m-1 \right){{e}^{2m}}-\left( m-1 \right)<0\Leftrightarrow \left( m-1 \right)\left( {{e}^{2m}}-1 \right)<0\)

\( \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m - 1 < 0\\
{e^{2m}} - 1 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m - 1 > 0\\
{e^{2m}} - 1 < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m < 1\\
m > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m > 1\\
m < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow 0 < m < 1 \Rightarrow \left\{ \begin{array}{l}
a = 0\\
b = 1
\end{array} \right..\)

Vậy \(P=a-3b=0-3=-\,3\).

Đặt $F\left( x \right) = {e^{2x}}\left( {a\cos 3x + b\sin 3x} \right) + c$.

Ta có

$F'\left( x \right) = 2a{e^{2x}}\cos 3x - 3a{e^{2x}}\sin 3x + 2b{e^{2x}}\sin 3x + 3b{e^{2x}}\cos 3x$ $ = \left( {2a + 3b} \right){e^{2x}}\cos 3x + \left( {2b - 3a} \right){e^{2x}}\sin 3x$

Để $F\left( x \right)$ là một nguyên hàm của hàm số $f\left( x \right) = {e^{2x}}\cos 3x$, điều kiện là

$F'\left( x \right) = {e^{2x}}\cos 3x \Leftrightarrow \left\{ \begin{array}{l}2a + 3b = 1\\2b - 3a = 0\end{array} \right.$ $ \Leftrightarrow \left\{ \begin{array}{l}a = \dfrac{2}{{13}}\\b = \dfrac{3}{{13}}\end{array} \right. \Rightarrow a + b = \dfrac{5}{{13}}$

Do đó \(\dfrac{1}{3} < S < \dfrac{1}{2}\).

Đặt \({{x}^{2}}+9=t\Rightarrow 2xdx=dt\Rightarrow xdx=\frac{1}{2}dt\).

Đổi cận:

$\begin{array}{l}
x = 0 \Rightarrow t = 9\\
x = 4 \Rightarrow t = 25
\end{array}$

Khi đó, ta có: \(I=\int\limits_{0}^{4}{x\ln ({{x}^{2}}+9)dx=}\frac{1}{2}\int\limits_{9}^{25}{\ln tdt}=\frac{1}{2}\left[ \left. t.\ln \left| t \right| \right|_{9}^{25}-\int_{9}^{25}{td(\ln t)} \right]=\frac{1}{2}\left[ t.\ln \left. t \right|_{9}^{25}-\int_{9}^{25}{t.\frac{1}{t}dt} \right]\)

\(=\frac{1}{2}\left[ t.\ln \left. t \right|_{9}^{25}-\int_{9}^{25}{dt} \right]=\frac{1}{2}\left[ t.\ln \left. t \right|_{9}^{25}-\left. t \right|_{9}^{25} \right]=\frac{1}{2}\left[ \left( 25\ln 25-9\ln 9 \right)-(25-9) \right]=25\ln 5-9\ln 3-8\)

Suy ra, \(a=25,\,b=-9,\,c=-8\Rightarrow T=a+b+c=8\)

Đặt $\left\{ \begin{array}{l}u = \ln \left( {x + 2} \right)\\{\rm{d}}v = x\,{\rm{d}}x\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\rm{d}}u = \dfrac{{{\rm{d}}x}}{{x + 2}}\\v = \dfrac{{{x^2}}}{2}\end{array} \right.,$ khi đó $I = \left. {\dfrac{{{x^2}\ln \left( {x + 2} \right)}}{2}} \right|_0^1 - \dfrac{1}{2}\int\limits_0^1 {\dfrac{{{x^2}}}{{x + 2}}{\rm{d}}x} .$

\(\begin{array}{l}\int\limits_0^2 {x\ln {{\left( {x + 1} \right)}^{2017}}dx}  = 2017\int\limits_0^2 {\ln \left( {x + 1} \right)d\left( {\frac{{{x^2}}}{2}} \right)} \\ = 2017\left[ {\left. {\ln \left( {x + 1} \right).\frac{{{x^2}}}{2}} \right|_0^2 - \int\limits_0^2 {\frac{{{x^2}}}{2}.\frac{1}{{x + 1}}dx} } \right] = 2017\left( {2\ln 3 - \frac{1}{2}\int\limits_0^2 {\left( {x - 1 + \frac{1}{{x + 1}}} \right)dx} } \right)\\ = 2017\left( {2\ln 3 - \frac{1}{2}\left. {\left( {\frac{{{x^2}}}{2} - x + \ln \left| {x + 1} \right|} \right)} \right|_0^2} \right)\\ = 2017\left( {2\ln 3 - \frac{1}{2}\ln 3} \right)\\ = 2017.\frac{3}{2}\ln 3 = \frac{{6051}}{2}\ln 3\\ \Rightarrow \left\{ \begin{array}{l}a = 6051\\b = 2\end{array} \right. \Rightarrow S = a - b = 6049\end{array}\)

\(\begin{align}  & f'\left( x \right)=-3.\frac{a}{{{\left( x+1 \right)}^{4}}}+b{{e}^{x}}+bx{{e}^{x}} \\  & \Rightarrow f'\left( 0 \right)=-3a+b=-22\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right) \\  & \int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( \frac{a}{{{\left( x+1 \right)}^{3}}}+bx{{e}^{x}} \right)dx}=a\int\limits_{0}^{1}{{{\left( x+1 \right)}^{-3}}dx}+b\int\limits_{0}^{1}{x{{e}^{x}}dx}=a{{I}_{1}}+b{{I}_{2}} \\  & {{I}_{1}}=\int\limits_{0}^{1}{{{\left( x+1 \right)}^{-3}}dx}=\left. \frac{{{\left( x+1 \right)}^{-2}}}{-2} \right|_{0}^{1}=\frac{-1}{2}\left( \frac{1}{4}-1 \right)=\frac{3}{8} \\ \end{align}\)

Đặt\(\left\{ \begin{array}{l}
u = x\\
dv = {e^x}dx
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
du = dx\\
v = {e^x}
\end{array} \right. \Rightarrow {I_2} = \left. {x{e^x}} \right|_0^1 - \int\limits_0^1 {{e^x}dx} = e - \left. {{e^x}} \right|_0^1 = e - \left( {e - 1} \right) = 1\)

\(\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=\frac{3}{8}a+b=5\,\,\,\,\,\,\left( 2 \right)\) 

Từ (1) và (2) \(\Rightarrow \left\{ \begin{align}  & a=8 \\  & b=2 \\ \end{align} \right.\)

\(I = \int\limits_0^1 {x\cos 2xdx} \) đặt \(\left\{ \begin{array}{l}
u = x\\
dv = \cos 2xdx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = dx\\
v = \frac{{\sin 2x}}{2}
\end{array} \right. \Rightarrow I = \left. {x.\frac{{\sin 2x}}{2}} \right|_0^1 - \frac{1}{2}\int\limits_0^1 {\sin 2xdx} \)

\(\begin{array}{l}
I = \frac{{\sin 2}}{2} + \frac{1}{2}.\left. {\frac{{\cos 2x}}{2}} \right|_0^1 = \frac{{\sin 2}}{2} + \frac{1}{4}\left( {\cos 2 - 1} \right) = \frac{1}{4}\left( {2\sin 2 + \cos 2 - 1} \right) = \frac{1}{4}\left( {a\sin 2 + b\cos 2 + c} \right)\\
\Rightarrow \left\{ \begin{array}{l}
a = 2\\
b = 1\\
c = - 1
\end{array} \right. \Rightarrow a - b + c = 0
\end{array}\)

$\int\limits_{0}^{2}{\left( 1-2x \right)f'\left( x \right)dx}=\int\limits_{0}^{2}{\left( 1-2x \right)d\left( f\left( x \right) \right)}=\left. \left( 1-2x \right)f\left( x \right) \right|_{0}^{2}+2\int\limits_{0}^{2}{f\left( x \right)dx} $

$\Leftrightarrow 2016=-3f\left( 2 \right)-f\left( 0 \right)+2\int\limits_{0}^{2}{f\left( x \right)dx}$

$\Leftrightarrow \int\limits_{0}^{2}{f\left( x \right)dx}=2016\left( 1 \right)$

Đặt \(x=2t\Leftrightarrow dx=2dt\), khi đó \(\left( 1 \right)\Leftrightarrow 2016=\int\limits_{0}^{1}{f\left( 2t \right)2f\left( t \right)}=2016\Leftrightarrow \int\limits_{0}^{1}{f\left( 2x \right)dx}=1008\)

\(\begin{array}{l}I = \int\limits_1^e {x\ln xdx}  = \int\limits_1^e {\ln xd\left( {\frac{{{x^2}}}{2}} \right)}  = \left. {\frac{{{x^2}}}{2}\ln x} \right|_1^e - \int\limits_1^e {\frac{{{x^2}}}{2}.\frac{{dx}}{x}}  = \frac{{{e^2}}}{2} - \frac{1}{2}\int\limits_1^e {xdx}  = \frac{{{e^2}}}{2} - \frac{1}{2}\left. {\frac{{{x^2}}}{2}} \right|_1^e = \frac{{{e^2}}}{2} - \frac{1}{4}\left( {{e^2} - 1} \right) = \frac{{{e^2} + 1}}{4}\\ \Rightarrow \left\{ \begin{array}{l}a = 1\\b = 1\\c = 4\end{array} \right. \Rightarrow T = a + b + c = 6\end{array}\)

Đặt  \(\left\{ \begin{array}{l}
u = \ln x\\
dv = xdx
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
du = \frac{{dx}}{x}\\
v = \frac{{{x^2}}}{2}
\end{array} \right.\)

\(\Rightarrow I=\left. \frac{{{x}^{2}}}{2}.\ln x \right|_{1}^{e}-\frac{1}{2}\int\limits_{1}^{e}{xdx}=\left. \frac{{{x}^{2}}}{2}.\ln x \right|_{1}^{e}-\left. \frac{{{x}^{2}}}{4} \right|_{1}^{e}=\frac{{{e}^{2}}}{2}-\left( \frac{{{e}^{2}}}{4}-\frac{1}{4} \right)=\frac{{{e}^{2}}+1}{4}\)

\(\Rightarrow a=b=\frac{1}{4}\Rightarrow a+b=\frac{1}{2}\)

Đặt\(\left\{ \begin{array}{l}
u = \ln \left( {1 + x} \right)\\
{\rm{d}}v = {\rm{d}}x
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{\rm{d}}u = \frac{{{\rm{d}}x}}{{x + 1}}\\
v = x
\end{array} \right.,\) khi đó \(I=\left. x.\ln \left( 1+x \right) \right|_{1}^{2}-\int\limits_{1}^{2}{\frac{x\text{d}x}{x+1}}=2.\ln 3-\ln 2-\int\limits_{1}^{2}{\frac{x}{x+1}\text{d}x}.\)

Ta có \(\int\limits_{1}^{2}{\frac{x}{x+1}\text{d}x}=\int\limits_{1}^{2}{\frac{x+1-1}{x+1}\text{d}x}=\int\limits_{1}^{2}{\left( 1-\frac{1}{x+1} \right)\text{d}x}=\left. \left( x-\ln \left| x+1 \right| \right) \right|_{1}^{2}=2-\ln 3-1+\ln 2=1+\ln 2-\ln 3\)

Vậy \(I=2.\ln 3-\ln 2-\left( 1+\ln 2-\ln 3 \right)=3.\ln 3-2.\ln 2-1=\ln \frac{27}{4}-1.\)

\(3f(x)+f'(x)=\sqrt{1+3{{e}^{-2x}}}\Leftrightarrow 3{{e}^{3x}}f(x)+{{e}^{3x}}f'(x)={{e}^{3x}}\sqrt{1+3{{e}^{-2x}}}\Leftrightarrow \left[ {{e}^{3x}}f(x) \right]'={{e}^{3x}}\sqrt{1+3{{e}^{-2x}}}\)

\(\Rightarrow \int\limits_{0}^{\frac{1}{2}\ln 6}{\left[ {{e}^{3x}}f(x) \right]'dx}=\int\limits_{0}^{\frac{1}{2}\ln 6}{{{e}^{3x}}\sqrt{1+3{{e}^{-2x}}}}dx\,\)

Ta có: \(\int\limits_{0}^{\frac{1}{2}\ln 6}{\left[ {{e}^{3x}}f(x) \right]'dx}=\left. \left( {{e}^{3x}}f(x) \right) \right|_{0}^{\frac{1}{2}\ln 6}={{e}^{\frac{3\ln 6}{2}}}f\left( \frac{1}{2}\ln 6 \right)-f(0)={{e}^{\ln \sqrt{{{6}^{3}}}}}f\left( \frac{1}{2}\ln 6 \right)-\frac{11}{3}=6\sqrt{6}.f\left( \frac{1}{2}\ln 6 \right)-\frac{11}{3}\)

\(\begin{align}  I=\int\limits_{0}^{\frac{1}{2}\ln 6}{{{e}^{3x}}\sqrt{1+3{{e}^{-2x}}}}dx=\int\limits_{0}^{\frac{1}{2}\ln 6}{{{e}^{2x}}\sqrt{{{e}^{2x}}+3}}dx=\frac{1}{2}\int\limits_{0}^{\frac{1}{2}\ln 6}{\sqrt{{{e}^{2x}}+3}}\,d\left( {{e}^{2x}}+3 \right) \\  =\frac{1}{2}\left. .\frac{{{\left( \sqrt{{{e}^{2x}}+3} \right)}^{3}}}{\frac{3}{2}} \right|_{0}^{\frac{1}{2}\ln 6}=\left. \frac{\left( {{e}^{2x}}+3 \right)\sqrt{{{e}^{2x}}+3}}{3} \right|_{0}^{\frac{1}{2}\ln 6}=9-\frac{8}{3}=\frac{19}{3} \\ \Rightarrow 6\sqrt{6}.f\left( \frac{1}{2}\ln 6 \right)-\frac{11}{3}=\frac{19}{3}\Rightarrow f\left( \frac{1}{2}\ln 6 \right)=\frac{10}{6\sqrt{6}}=\frac{5\sqrt{6}}{18} \\ \end{align}\)

Đặt \(t=\sqrt{3x+1}\Leftrightarrow {{t}^{2}}=3x+1\Leftrightarrow 2tdt=3dx\)

Đổi cận \(\left\{ \begin{align}   x=0\Leftrightarrow t=1 \\   x=1\Leftrightarrow t=2 \\ \end{align} \right.\), khi đó ta có:  \(\int\limits_{0}^{1}{3{{e}^{\sqrt{3x+1}}}dx}=\int\limits_{1}^{2}{{{e}^{t}}.2tdt}=2\int\limits_{1}^{2}{t{{e}^{t}}dt}\)

Đặt \(\left\{ \begin{array}{l}u = t\\dv = {e^t}dt\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}du = dt\\v = {e^t}\end{array} \right. \Rightarrow 2\int\limits_1^2 {t{e^t}dt}  = 2\left( {\left. {t{e^t}} \right|_1^2 - \int\limits_1^2 {{e^t}dt} } \right) = 2\left( {\left. {t{e^t}} \right|_1^2 - \left. {{e^t}} \right|_1^2} \right) = 2\left( {2{e^2} - e - \left( {{e^2} - e} \right)} \right) = 2{e^2}\)

\(\Rightarrow \left\{ \begin{align}  a=10 \\   b=c=0 \\ \end{align} \right.\Rightarrow P=a+b+c=10\)

$\int\limits_{1}^{e}{\frac{f(x)}{x}dx}=\int\limits_{1}^{e}{f(x)d\ln x}=\left. f(x)\ln x \right|_{1}^{e}-\int\limits_{1}^{e}{\ln xf'(x)dx}=1$

$\Rightarrow f(e)-\int\limits_{1}^{e}{\ln xf'(x)dx}=1$

$\Leftrightarrow \int\limits_{1}^{e}{\ln xf'(x)dx}=f(e)-1=2-1=1$

Vì ${x^2}$ là một nguyên hàm của hàm số $f\left( x \right){e^{2x}} \Rightarrow \int {f\left( x \right){e^{2x}}\,{\rm{d}}x}  = {x^2}.$

Đặt $\left\{ \begin{array}{l}u = {e^{2x}}\\{\rm{d}}v = f'\left( x \right){\rm{d}}x\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\rm{d}}u = 2{e^{2x}}{\rm{d}}x\\v = f\left( x \right)\end{array} \right.,$ khi đó $\int\limits_0^1 {f'\left( x \right){e^{2x}}{\rm{d}}x}  = \left. {f\left( x \right){e^{2x}}} \right|_0^1 - 2\int\limits_0^1 {f\left( x \right){e^{2x}}\,{\rm{d}}x} .$

Suy ra $I = {e^2}f\left( 1 \right) - f\left( 0 \right) - 2\left. {{x^2}} \right|_0^1 =  2-0 - 2 =  0$

Vậy $I =  0$

Đặt $\left\{ \begin{array}{l}u = x + \ln x\\{\rm{d}}v = \dfrac{{{\rm{d}}x}}{{{{\left( {x + 1} \right)}^3}}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\rm{d}}u = \dfrac{{x + 1}}{x}\,{\rm{d}}x\\v =  - \dfrac{1}{{2{{\left( {x + 1} \right)}^2}}}\end{array} \right.$.

Khi đó $I = \left. { - \dfrac{{x + \ln x}}{{2{{\left( {x + 1} \right)}^2}}}} \right|_1^2 + \int\limits_1^2 {\dfrac{{x + 1}}{x}.\dfrac{1}{{2{{\left( {x + 1} \right)}^2}}}{\rm{d}}x} .$

$ =  - \dfrac{{2 + \ln 2}}{{18}} + \dfrac{1}{8} + \dfrac{1}{2}\int\limits_1^2 {\dfrac{{{\rm{d}}x}}{{x\left( {x + 1} \right)}}} $$=  - \dfrac{{2 + \ln 2}}{{18}} + \dfrac{1}{8} + \dfrac{1}{2}\int\limits_1^2 {\left( {\dfrac{1}{x} - \dfrac{1}{{x + 1}}} \right){\rm{d}}x} .$

$\begin{array}{l} =  - \dfrac{{2 + \ln 2}}{{18}} + \dfrac{1}{8} + \dfrac{1}{2}\left. {\left( {\ln \left| x \right| - \ln \left| {x + 1} \right|} \right)} \right|_1^2\\ = \dfrac{1}{{72}} - \dfrac{1}{{18}}\ln 2 + \dfrac{1}{2}\left( {\ln 2 - \ln 3 + \ln 2} \right)\\ = \dfrac{1}{{72}} + \dfrac{{17}}{{18}}\ln 2 - \dfrac{1}{2}\ln 3\\ = a + b.\ln 2 - c.\ln 3.\end{array}$

Vậy $\left\{ \begin{array}{l}a = \dfrac{1}{{72}}\\b = \dfrac{{17}}{{18}}\\c = \dfrac{1}{2}\end{array} \right. \Rightarrow \,\dfrac{c}{a} = \dfrac{1}{2}:\dfrac{1}{{72}} = 36.$

Đặt $\left\{ \begin{array}{l}u = 1 - \ln x\\dv = 2xdx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du =  - \dfrac{{dx}}{x}\\v = {x^2}\end{array} \right.$

$I = {x^2}\left( {1 - \ln x} \right)\left| {\begin{array}{*{20}{c}}{^e}\\{_1}\end{array}} \right. - \int\limits_1^e { - xdx}  =  - 1 + \dfrac{{{x^2}}}{2}\left| {\begin{array}{*{20}{c}}{^e}\\{_1}\end{array}} \right. =  - 1 + \left( {\dfrac{{{e^2}}}{2} - \dfrac{1}{2}} \right) = \dfrac{{{e^2} - 3}}{2}$

Dùng máy tính kiểm tra từng đáp án hoặc:

Đặt $u = \ln x,dv = xdx \Rightarrow du = \dfrac{{dx}}{x},v = \dfrac{{{x^2}}}{2}$

$I = \dfrac{{{x^2}\ln x}}{2}\left| {\begin{array}{*{20}{c}}{^e}\\{_1}\end{array}} \right. - \int\limits_1^e {\dfrac{x}{2}dx}  = \dfrac{{{e^2}}}{2} - \dfrac{{{x^2}}}{4}\left| {\begin{array}{*{20}{c}}{^e}\\{_1}\end{array}} \right. = \dfrac{{{e^2}}}{2} - \left( {\dfrac{{{e^2}}}{4} - \dfrac{1}{4}} \right) = \dfrac{{{e^2} + 1}}{4}$

Đặt  $\left\{ \begin{array}{l}u = \ln x\\dv = \dfrac{{dx}}{{{{(x + 1)}^2}}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \dfrac{{dx}}{x}\\v =  - \dfrac{1}{{x + 1}}\end{array} \right.$

$\begin{array}{l} \Rightarrow I =  - \dfrac{{\ln x}}{{x + 1}}\left| {_1^{{2^{1000}}} + \int\limits_1^{{2^{1000}}} {\dfrac{1}{{x + 1}}.} \dfrac{{dx}}{x} = } \right. - \dfrac{{\ln {2^{1000}}}}{{{2^{1000}} + 1}} + \int\limits_1^{{2^{1000}}} {\left( {\dfrac{1}{x} - \dfrac{1}{{x + 1}}} \right)dx}  \\ =  - \dfrac{{1000\ln 2}}{{{2^{1000}} + 1}} + \left. {\ln \left| {\dfrac{x}{{x + 1}}} \right|} \right|_1^{{2^{1000}}} =  - \dfrac{{1000\ln 2}}{{{2^{1000}} + 1}} + \ln \dfrac{{{2^{1000}}}}{{{2^{1000}} + 1}} - \ln \dfrac{1}{2} \\ =  - \dfrac{{1000\ln 2}}{{{2^{1000}} + 1}} + \ln \dfrac{{{2^{1001}}}}{{{2^{1000}} + 1}}\end{array}$

Đặt$f\left( x \right) = {e^{2x}}\left( {a\cos 3x + b\sin 3x} \right) + c$. Ta có

$f'\left( x \right) = 2a{e^{2x}}\cos 3x - 3a{e^{2x}}\sin 3x + 2b{e^{2x}}\sin 3x + 3b{e^{2x}}\cos 3x $

$= \left( {2a + 3b} \right){e^{2x}}\cos 3x + \left( {2b - 3a} \right){e^{2x}}\sin 3x$

Để $f\left( x \right)$ là một nguyên hàm của hàm số ${e^{2x}}\cos 3x$, điều kiện là

$f'\left( x \right) = {e^{2x}}\cos 3x \Leftrightarrow \left\{ \begin{array}{l}2a + 3b = 1\\2b - 3a = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = \dfrac{2}{{13}}\\b = \dfrac{3}{{13}}\end{array} \right. \Rightarrow a + b = \dfrac{5}{{13}}$