What is the integral of ((e^{2x}-e^x)^2)/(e^{4x)} ?

The integral of ((e^{2x}-e^x)^2)/(e^{4x)} is x+2/(e^x)-1/2 e^{-2x}+C

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integral of ((e^{2x}-e^x)^2)/(e^{4x)}

\int \frac{( e ^{2 x} - e ^{x} ) ^{2}}{e ^{4 x}} d x

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

Solution steps

\int \frac{( e ^{2 x} - e ^{x} ) ^{2}}{e ^{4 x}} d x

Expand

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

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

\int 1 d x = x

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

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

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

Simplify

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

Add a constant to the solution

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

What is the integral of ((e^{2x}-e^x)^2)/(e^{4x)} ?
The integral of ((e^{2x}-e^x)^2)/(e^{4x)} is x+2/(e^x)-1/2 e^{-2x}+C