What is expand 1+((e^x-e^{-x})/2)^2 ?

The solution to expand 1+((e^x-e^{-x})/2)^2 is 1+(e^{2x})/4-1/2+(e^{-2x})/4

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expand 1+((e^x-e^{-x})/2)^2

expand

1 + (\frac{e ^{x} - e ^{- x}}{2})^{2}

1 + \frac{e ^{2 x}}{4} - \frac{1}{2} + \frac{e ^{- 2 x}}{4}

Solution steps

1 + (\frac{e ^{x} - e ^{- x}}{2})^{2}

(\frac{e ^{x} - e ^{- x}}{2})^{2} = \frac{e ^{2 x} - 2 + e ^{- 2 x}}{2 ^{2}}

= 1 + \frac{e ^{2 x} + e ^{- 2 x} - 2}{2 ^{2}}

2^{2} = 4

= 1 + \frac{e ^{2 x} + e ^{- 2 x} - 2}{4}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{e ^{2 x} - 2 + e ^{- 2 x}}{4} = \frac{e ^{2 x}}{4} - \frac{2}{4} + \frac{e ^{- 2 x}}{4}

= 1 + \frac{e ^{2 x}}{4} - \frac{2}{4} + \frac{e ^{- 2 x}}{4}

Cancel

\frac{2}{4} : \frac{1}{2}

= 1 + \frac{e ^{2 x}}{4} - \frac{1}{2} + \frac{e ^{- 2 x}}{4}

\frac{1}{x ( x ^{2} + x + 1 )}

(x^{2} + 2 x + 1)^{6}

\frac{( x ^{4} - 2 x ^{3} + 5 x ^{2} - 3 x + 4 )}{( x - 1 ) ( x + 2 )}

\frac{1}{2} (e^{x^{2}} + e^{- x^{2}})

- \frac{x ^{2} + 6 x + 21}{( x ^{2} - 3 ) ( x + 4 )}

What is expand 1+((e^x-e^{-x})/2)^2 ?
The solution to expand 1+((e^x-e^{-x})/2)^2 is 1+(e^{2x})/4-1/2+(e^{-2x})/4