[tex]f(x) = x+1+\ln x\\ \\ g\Big(f(x)\Big) = x \Rightarrow g(x+1+\ln x) = x \\ \\ \displaystyle I =\int_{2}^{e+2} g(x)\, dx\\ \\ x = t+1+\ln t \Rightarrow dx = \Big(1+\dfrac{1}{t}\Big)\, dt \\ x = 2 \Rightarrow t = 1\\ x = e+2 \Rightarrow t = e \\ \\ I = \int_{1}^{e} g(t+1+\ln t)\cdot \Big(1+\dfrac{1}{t}\Big)\, dt =\\ \\ =\int_{1}^{e}t\cdot \Big(1+\dfrac{1}{t}\Big)\,dt=\int_{1}^{e}(t+1)\,dt = \Big(\dfrac{t^2}{2}+t\Big)\Bigg|_{1}^e =\\ \\ = \dfrac{e^2}{2}+e - \dfrac{1}{2} - 1= \dfrac{1}{2}\Big(e^2+2e-3\Big)[/tex]
