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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!

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    $\begingroup$ Square $e^x+e^{-x}$ to see what's going on. $\endgroup$ – André Nicolas Jan 12 '16 at 0:47
  • $\begingroup$ @AndréNicolas ah, I see. Thank you! $\endgroup$ – JayJensen Jan 12 '16 at 0:51
  • $\begingroup$ You are welcome. This is probably not the last time you will bump into something similar. $\endgroup$ – André Nicolas Jan 12 '16 at 0:56
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We have $$\begin{align}\int_{0}^{1} \sqrt{e^{2x}+e^{-2x}+2}\,dx &= \int_0^1\sqrt{{e^{2x} \over 1} + {1 \over e^{2x}} + {2 \over 1}}\,dx\\ & =\int_0^1\sqrt{{e^{4x} + 1 + 2e^{2x} \over e^{2x}}} \,dx \\ &=\int_0^1{\sqrt{e^{4x} + 2e^{2x} + 1} \over e^{x}}\,dx \\ &= \int_0^1{\sqrt{(e^{2x}+1)^2} \over e^{x}}\,dx \\& =\int_0^1{e^{2x} + 1 \over e^x}\,dx \\ &=\int_0^1\left[{e^{2x} \over e^x} + {1 \over e^x}\right]\,dx \\ &=\int_0^1(e^x + e^{-x})\,dx\\ &=\left.e^x - e^{-x}\right]_0^1 \\ &=e - {1 \over e}.\end{align}$$

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  • $\begingroup$ You make an easy calculation complicated.... $\endgroup$ – Empty Jan 12 '16 at 15:27
  • $\begingroup$ @decaf-math Where did you get that picture of me $\endgroup$ – 2316354654 Sep 29 '17 at 5:03
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Hint : $\displaystyle e^{2x}+e^{-2x}+2=(e^x)^2+2.e^x.e^{-x}+(e^{-x})^2=(e^x+e^{-x})^2$.

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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,$$\int\limits_{0}^{1}\sqrt{e^{2x}+e^{-2x}+2}\text{d}x=\int\limits_{0}^{1}\sqrt{4\cosh^2(x)}\text{d}x=2\int\limits_{0}^{1}\cosh(x)\text{d}x=2\Big.\sinh(x)\Big\vert_{0}^{1}=2\sinh(1)$$

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