Biết $F\left( x \right) = \left( {ax + b} \right).{e^x}$ là nguyên hàm của hàm số $y = \left( {2x + 3} \right).{e^x}$. Khi đó $b - a$ là

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

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

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

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

 \(\frac{61}{9}\)

 $4.$

$61.$

 \(\frac{61}{3}\).

$\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} $

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

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

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

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

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

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

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

\(\sqrt x \sin \sqrt x  + \cos \sqrt x  + C\)                  

\(\sqrt x \sin \sqrt x  - \cos \sqrt x  + C\)

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

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

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

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