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

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

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simplify (1+e^{2x})e^{-x}dx

simplify

(1 + e^{2 x}) e^{- x} d x

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

Solution steps

(1 + e^{2 x}) e^{- x} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

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

Multiply fractions:

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

= \frac{1 \cdot ( 1 + e ^{2 x} ) d x}{e ^{x}}

Multiply:

1 \cdot (1 + e^{2 x}) = (1 + e^{2 x})

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

e^{- 0.208}

7 0, 2 x^{0.28} 0.2

e^{- 31}

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

e^{- (- 0)}

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