What is the integral of (5-e^{-x})/(e^{-x)} ?

The integral of (5-e^{-x})/(e^{-x)} is 5e^x-x+C

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integral of (5-e^{-x})/(e^{-x)}

\int \frac{5 - e ^{- x}}{e ^{- x}} d x

5 e^{x} - x + C

Solution steps

\int \frac{5 - e ^{- x}}{e ^{- x}} d x

Expand

\frac{5 - e ^{- x}}{e ^{- x}} : \frac{5}{e ^{- x}} - \frac{e ^{- x}}{e ^{- 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{5}{e ^{- x}} d x - \int \frac{e ^{- x}}{e ^{- x}} d x

\int \frac{5}{e ^{- x}} d x = 5 e^{x}

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

= 5 e^{x} - x

Add a constant to the solution

= 5 e^{x} - x + C

\int \frac{x ^{2}}{x - 3} d x

\int \frac{6 x}{4 x ^{2} + 4 x + 10} d x

\frac{d}{d x} (e^{s i n (x - 1)})

\int_{2}^{5} x^{6} d x

x y^{^{'}} - 2 y^{^{'}} - (x - 2) cos (x) + y = 0

What is the integral of (5-e^{-x})/(e^{-x)} ?
The integral of (5-e^{-x})/(e^{-x)} is 5e^x-x+C