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

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

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

expand

e^{x} (e^{1 + x} - e^{- 1 - x})

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

Solution steps

e^{x} (e^{1 + x} - e^{- 1 - x})

Apply the distributive law:

a (b - c) = ab - a c

a = e^{x}, b = e^{1 + x}, c = e^{- 1 - x}

= e^{x} e^{1 + x} - e^{x} e^{- 1 - x}

Simplify

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

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

\frac{1}{60} \cdot 8 - \frac{1}{4} \cdot 4 + \frac{37}{30} \cdot 2

((2 x^{2} y - 2 y^{2} + 2 x y) d x + (x^{2} - 2 y) d y = 0) e^{2 x}

\frac{( x - 9 ) ( x + 1 ) ( x + 6 )}{( x + 8 ) ( x + 4 )}

(a - b 2) (a + b 2)

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