Đặt \(t=3x+4\Rightarrow x=\frac{t-4}{3} ;{\rm d}x=\frac{{\rm d}t}{3} \)
\(I=\frac{1}{9} \int t^{10} \left(t-4\right){\rm d}t =\frac{1}{9} \int \left(t^{11} -4t^{10} \right){\rm d}t =\frac{1}{108} t^{12} -\frac{4}{99} t^{11} +C\)
\(=\frac{1}{108} \left(3x+4\right)^{12} -\frac{4}{99} \left(3x+4\right)^{11} +C.\)