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

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

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

\int \frac{e ^{x} - 4 e ^{- x}}{e ^{x}} d x

x + 2 e^{- 2 x} + C

Solution steps

\int \frac{e ^{x} - 4 e ^{- x}}{e ^{x}} d x

Expand

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

= \int 1 - \frac{4}{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{4}{e ^{2 x}} d x

\int 1 d x = x

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

= x - (- 2 e^{- 2 x})

Simplify

= x + 2 e^{- 2 x}

Add a constant to the solution

= x + 2 e^{- 2 x} + C

y^{3} - (10 x + 2) + 3 x y^{2} \frac{d y}{d x} = 0

\frac{\partial}{\partial x} (x e^{y} + y + 1)

\frac{\partial}{\partial y} (\frac{x ^{2} - y ^{2}}{2})

\int_{0}^{0.02} 200 x d x

\frac{3 s + 1}{s ( s + 2 )}

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