Cho $I = \int {x\sqrt {3{x^2} + 1} dx}  = \dfrac{1}{a}\sqrt {{{\left( {3{x^2} + 1} \right)}^b}}  + C$. Giá trị $a$ và $b$ lần lượt là:

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

\(dt = \dfrac{{v'\left( x \right)}}{{u'\left( t \right)}}dx\)      

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

\(dx = \dfrac{{v'\left( x \right)}}{{u'\left( t \right)}}dt\)

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

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

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

\(\int {2xf\left( {{x^2}} \right)dx}  = 2\int {tf\left( t \right)dt} \)

\(\dfrac{{12}}{{79}}\)

\(\dfrac{{95}}{{103}}\)

\(\dfrac{{22}}{{105}}\)

\(\dfrac{{48}}{{109}}\)