Trong phương pháp nguyên hàm từng phần, nếu \(\left\{ \begin{array}{l}u = g\left( x \right)\\dv = h\left( x \right)dx\end{array} \right.\) thì:

\(\int {udv}  = uv + \int {vdu} \)

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

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

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

\(I = \left( {x + 1} \right)f\left( x \right) - 2F\left( x \right) + C\)     

\(I = F\left( x \right) - \left( {x + 1} \right)f\left( x \right)\)

$I = \left( {x + 1} \right)f\left( x \right) + C$

\(I = \left( {x + 1} \right)f\left( x \right) - F\left( x \right) + C\)  

$\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$  

$3$

$1$

$2$