What is the integral of ((e^{-x}+6)^2)/(e^x) ?

The integral of ((e^{-x}+6)^2)/(e^x) is -1/3 e^{-3x}-6e^{-2x}-(36)/(e^x)+C

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integral of ((e^{-x}+6)^2)/(e^x)

\int \frac{( e ^{- x} + 6 ) ^{2}}{e ^{x}} d x

- \frac{1}{3} e^{- 3 x} - 6 e^{- 2 x} - \frac{36}{e ^{x}} + C

Solution steps

\int \frac{( e ^{- x} + 6 ) ^{2}}{e ^{x}} d x

Expand

\frac{( e ^{- x} + 6 ) ^{2}}{e ^{x}} : \frac{1}{e ^{3 x}} + \frac{12}{e ^{2 x}} + \frac{36}{e ^{x}}

= \int \frac{1}{e ^{3 x}} + \frac{12}{e ^{2 x}} + \frac{36}{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{12}{e ^{2 x}} d x + \int \frac{36}{e ^{x}} d x

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

\int \frac{12}{e ^{2 x}} d x = - 6 e^{- 2 x}

\int \frac{36}{e ^{x}} d x = - \frac{36}{e ^{x}}

= - \frac{1}{3} e^{- 3 x} - 6 e^{- 2 x} - \frac{36}{e ^{x}}

Add a constant to the solution

= - \frac{1}{3} e^{- 3 x} - 6 e^{- 2 x} - \frac{36}{e ^{x}} + C

\frac{d}{d x} (5^{x} ln (x))

(2 x y - 3) d x + (x^{2} + 4 y) d y = 0

\int_{0}^{l n (18)} x e^{x} d x

\frac{d}{d x} (e^{i x} + e^{- i x})

\int 6 e^{- 3 x} d x

What is the integral of ((e^{-x}+6)^2)/(e^x) ?
The integral of ((e^{-x}+6)^2)/(e^x) is -1/3 e^{-3x}-6e^{-2x}-(36)/(e^x)+C