Nếu \(t = {x^2}\) thì:

\(dt = u'\left( x \right)dx\)

\(dx = u'\left( t \right)dt\) 

\(dt = \dfrac{1}{{u\left( x \right)}}dx\)

\(dx = \dfrac{1}{{u\left( t \right)}}dt\) 

$\int {f\left( {3x} \right){\mkern 1mu} {\rm{d}}x = 2x\ln \left( {9x - 1} \right) + C.} $

$\int {f\left( {3x} \right){\mkern 1mu} {\rm{d}}x = 6x\ln \left( {3x - 1} \right) + C.} $

$\int {f\left( {3x} \right){\mkern 1mu} {\rm{d}}x = 6x\ln \left( {9x - 1} \right) + C.} $

$\int {f\left( {3x} \right){\mkern 1mu} {\rm{d}}x = 3x\ln \left( {9x - 1} \right) + C.} $

\(f\left( x \right)dx =  - tdt\)              

\(f\left( x \right)dx = 2tdt\)

\(f\left( x \right)dx =  - 2{t^2}dt\)     

\(f\left( x \right)dx = 2{t^2}dt\)

\(I = \dfrac{1}{5}{\left( {{x^3} + 1} \right)^2}\sqrt {{x^3} + 1}  - \dfrac{1}{3}\left( {{x^3} + 1} \right)\sqrt {{x^3} + 1}  + C\)

\(I = \dfrac{2}{5}{\left( {{x^3} + 1} \right)^2}\sqrt {{x^3} + 1}  - \dfrac{2}{3}\left( {{x^3} + 1} \right)\sqrt {{x^3} + 1}  + C\)

\(I = \dfrac{2}{5}{\left( {{x^3} + 1} \right)^2}\sqrt {{x^3} + 1}  + C\)                   

\(I = \dfrac{2}{5}{\left( {{x^3} + 1} \right)^2}\sqrt {{x^3} + 1}  + \left( {{x^3} + 1} \right)\sqrt {{x^3} + 1}  + C\)

\(F\left( x \right) =  - 2\sqrt {1 - \ln x}  + \dfrac{2}{3}\left( {1 - \ln x} \right)\sqrt {1 - \ln x}  + 3\)

\(F\left( x \right) =  - \sqrt {1 - \ln x}  + \dfrac{1}{3}\left( {1 - \ln x} \right)\sqrt {1 - \ln x}  + 3\)

\(F\left( x \right) =  - 2\sqrt {1 - \ln x}  - \dfrac{2}{3}\left( {1 - \ln x} \right)\sqrt {1 - \ln x}  + 3\)

\(F\left( x \right) = 2\sqrt {1 - \ln x}  - \dfrac{2}{3}\left( {1 - \ln x} \right)\sqrt {1 - \ln x}  + 3\)

\(I = t - \dfrac{{{t^2}}}{2} + C\)

\(I = \dfrac{{{t^2}}}{2} - t + C\)       

\(I = \dfrac{{{t^2}}}{2} - \dfrac{{{t^2}}}{3} + C\)

\(I =  - \dfrac{{{t^2}}}{2} + \dfrac{{{t^2}}}{3} + C\)