Cho \(F\left( x \right) = \int {\left( {x + 1} \right)f'\left( x \right)dx} \). Tính \(I = \int {f\left( x \right)dx} \) theo $F(x)$.

\(\int {udv}  = uv + \int {vdu} \)

\(\int {udv}  = uv - \int {vdu} \)          

\(\int {udv}  = \int {uv}  - \int {vdu} \)

\(\int {udv}  = \int {uvdv}  - \int {vdu} \)

\(\left\{ \begin{array}{l}du = g'\left( x \right)dx\\v = \int {h\left( x \right)dx} \end{array} \right.\)

\(\left\{ \begin{array}{l}du = g\left( x \right)dx\\v = \int {h\left( x \right)dx} \end{array} \right.\)           

\(\left\{ \begin{array}{l}du = \int {g\left( x \right)dx} \\v = h\left( x \right)\end{array} \right.\)  

\(\left\{ \begin{array}{l}du = g'\left( x \right)dx\\v = h\left( x \right)\end{array} \right.\)

$\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} $

\(\dfrac{1}{4}{x^4}\ln 3x + C\)         

\( - \dfrac{1}{4}{x^4}\ln 3x - \dfrac{1}{{16}}{x^4} + C\)

\( - \dfrac{1}{4}{x^4}\ln 3x + \dfrac{1}{{16}}{x^4} + C\)

\(\dfrac{1}{4}{x^4}\ln 3x - \dfrac{1}{{16}}{x^4} + C\)

$a + b = 2$  

$a + b = 3$                

$a + b = 0$     

$a + b = 1$

$ - 1$  

$3$

$1$

$2$