Biết rằng $F\left( x \right) = {e^{2x}}\left( {a\cos 3x + b\sin 3x} \right)$ là một nguyên hàm của hàm số $f\left( x \right) = {e^{2x}}\cos 3x$, trong đó a, b, c là các hằng số. Giá trị của tổng $S = a + b$ thỏa mãn:

$C.F\left( x \right)$

$C - F\left( x \right)$

$C + F\left( x \right)$

$\dfrac{{F\left( x \right)}}{C}$

 $\frac{61}{9}$

 $4.$

$61.$

 $\frac{61}{3}$.

$\int {f(x)dx = {x^3}\ln 3x - \dfrac{{{x^3}}}{3} + C}$        

$\int {f(x)dx = \dfrac{{{x^3}\ln 3x}}{3} - \dfrac{{{x^3}}}{9} + C}$

$\int {f(x)dx = \dfrac{{{x^3}\ln 3x}}{3} - \dfrac{{{x^3}}}{3} + C}$          

$\int {f(x)dx = \dfrac{{{x^3}\ln 3x}}{3} - \dfrac{{{x^3}}}{{27}} + C}$

$\int{\left( {{u}^{4}}+5{{u}^{3}} \right)du}$                       

$\int\limits_{ - 1}^2 {f\left( {2x} \right)} d{\rm{x  =  2}}$

$\int\limits_{ - 3}^3 {f\left( {x + 1} \right)} d{\rm{x  =  2}}$

$\int\limits_{ - 1}^2 {f\left( {2x} \right)} d{\rm{x  =  1}}$

$\int\limits_0^6 {\dfrac{1}{2}f\left( {x - 2} \right)} d{\rm{x  =  1}}$

$2\left( {\sqrt x \sin \sqrt x  - \cos \sqrt x } \right) + C$      

$2\left( {\sqrt x \sin \sqrt x  + \cos \sqrt x } \right) + C$

$\sqrt x \sin \sqrt x  + \cos \sqrt x  + C$                  

$\sqrt x \sin \sqrt x  - \cos \sqrt x  + C$

$I = \left. {\dfrac{{{x^2}\ln \left( {x + 2} \right)}}{2}} \right|_0^1 - \dfrac{1}{2}\int\limits_0^1 {\dfrac{{{x^2}}}{{x + 2}}{\rm{d}}x} .$

$I = \left. {{x^2}\ln \left( {x + 2} \right)} \right|_0^1 - \dfrac{1}{4}\int\limits_0^1 {\dfrac{{{x^2}}}{{x + 2}}{\rm{d}}x} .$

$I = \left. {\dfrac{{{x^2}\ln \left( {x + 2} \right)}}{2}} \right|_0^1 + \int\limits_0^1 {\dfrac{{{x^2}}}{{x + 2}}{\rm{d}}x} .$

$I = \left. {\dfrac{{{x^2}\ln \left( {x + 2} \right)}}{4}} \right|_0^1 - \dfrac{1}{4}\int\limits_0^1 {\dfrac{{{x^2}}}{{x + 2}}{\rm{d}}x} .$