What is the integral of ((e^{2x}-2e^{-x}+1))/(e^x) ?

The integral of ((e^{2x}-2e^{-x}+1))/(e^x) is e^x+e^{-2x}-1/(e^x)+C

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

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

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

Solution steps

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int e^{x} d x = e^{x}

\int \frac{2}{e ^{2 x}} d x = - e^{- 2 x}

\int \frac{1}{e ^{x}} d x = - \frac{1}{e ^{x}}

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

Simplify

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

Add a constant to the solution

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

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What is the integral of ((e^{2x}-2e^{-x}+1))/(e^x) ?
The integral of ((e^{2x}-2e^{-x}+1))/(e^x) is e^x+e^{-2x}-1/(e^x)+C