What is the integral of ((1+e^x)^2)/(e^{3x)} ?

The integral of ((1+e^x)^2)/(e^{3x)} is -1/3 e^{-3x}-e^{-2x}-1/(e^x)+C

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integral of ((1+e^x)^2)/(e^{3x)}

\int \frac{( 1 + e ^{x} ) ^{2}}{e ^{3 x}} d x

- \frac{1}{3} e^{- 3 x} - e^{- 2 x} - \frac{1}{e ^{x}} + C

Solution steps

\int \frac{( 1 + e ^{x} ) ^{2}}{e ^{3 x}} d x

Expand

\frac{( 1 + e ^{x} ) ^{2}}{e ^{3 x}} : \frac{1}{e ^{3 x}} + \frac{2}{e ^{2 x}} + \frac{1}{e ^{x}}

= \int \frac{1}{e ^{3 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 \frac{1}{e ^{3 x}} d x + \int \frac{2}{e ^{2 x}} d x + \int \frac{1}{e ^{x}} d x

\int \frac{1}{e ^{3 x}} d x = - \frac{1}{3} e^{- 3 x}

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

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

= - \frac{1}{3} e^{- 3 x} - e^{- 2 x} - \frac{1}{e ^{x}}

Add a constant to the solution

= - \frac{1}{3} e^{- 3 x} - e^{- 2 x} - \frac{1}{e ^{x}} + C

What is the integral of ((1+e^x)^2)/(e^{3x)} ?
The integral of ((1+e^x)^2)/(e^{3x)} is -1/3 e^{-3x}-e^{-2x}-1/(e^x)+C