$t=2x\\ dt=2dx \Rightarrow dx=\dfrac{dt}{2}\\ \text{Đổi cận}\\ x | \quad 1 \quad | \quad 2\\ \overline{t\, | \quad 2 \quad | \quad 4}\\ \displaystyle\int\limits^2_1 f(x) \, dx=8\\ \Leftrightarrow \dfrac{1}{2}\displaystyle\int\limits^4_2 f(t) \, dt=8\\ \Leftrightarrow \displaystyle\int\limits^4_2 f(t) \, dt=16\\ \Leftrightarrow \displaystyle\int\limits^4_2 f(x) \, dx=16\\ \displaystyle\int\limits^4_1 [2f(x)+1] \, dx\\ =2\displaystyle\int\limits^4_1 f(x) \, dx +\displaystyle\int\limits^4_1 dx\\ =2\left(\displaystyle\int\limits^2_1 f(x) \, dx +\displaystyle\int\limits^4_2 f(x) \, dx \right)+x \Big|^4_1\\ =2(3+16)+3\\ =41\\ \Rightarrow C$