Integral Problem: $\int_{0}^{1} \sqrt{e^{2x}+e^{-2x}+2}$ I am having trouble with this definite integral problem
$$\int_{0}^{1} \sqrt{e^{2x}+e^{-2x}+2} \, dx$$
I know that the solution is 
$$e - \dfrac{1}{e}$$
I checked the step by step solution from Wolfram Alpha and it says because $$ 0 < x < 1 $$ the integral can be simplified to:
$$\int_{0}^{1} (e^{-x} + e^{x}) \,dx $$
I don't quite understand this step or if Wolfram Alpha just left some vital steps out of the solution. Any help or links would be greatly appreciated!

Another option:$$e^{2x}+e^{-2x}+2=2\frac{e^{2x}+e^{-2x}}{2}+2=2\cosh(2x)+2=2\left[2\cosh^2(x)-1\right]+2=4\cosh^2(x)$$Hence,$$\int\limits_{0}^{1}\sqrt{e^{2x}+e^{-2x}+2}\text{d}x=\int\limits_{0}^{1}\sqrt{4\cosh^2(x)}\text{d}x=2\int\limits_{0}^{1}\cosh(x)\text{d}x=2\Big.\sinh(x)\Big\vert_{0}^{1}=2\sinh(1)$$
