What is the integral of (e^{-x}-e^5)(e^x+e^{-3}) ?

The integral of (e^{-x}-e^5)(e^x+e^{-3}) is x-e^{-x-3}-e^{5+x}-e^2x+C

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integral of (e^{-x}-e^5)(e^x+e^{-3})

\int (e^{- x} - e^{5}) (e^{x} + e^{- 3}) d x

x - e^{- x - 3} - e^{5 + x} - e^{2} x + C

Solution steps

\int (e^{- x} - e^{5}) (e^{x} + e^{- 3}) d x

Expand

(e^{- x} - e^{5}) (e^{x} + e^{- 3}) : 1 + e^{- x - 3} - e^{5 + x} - e^{2}

= \int 1 + e^{- x - 3} - e^{5 + x} - e^{2} 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 e^{- x - 3} d x - \int e^{5 + x} d x - \int e^{2} d x

\int 1 d x = x

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

\int e^{5 + x} d x = e^{5 + x}

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

= x - e^{- x - 3} - e^{5 + x} - e^{2} x

Add a constant to the solution

= x - e^{- x - 3} - e^{5 + x} - e^{2} x + C

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

\frac{d}{d x} (\frac{1}{3 x ^{2} + x + 5})

\int_{0}^{2} e^{x} - e^{- x} d x

s l o p e y = 3 x^{2} + 2 xx a

a re a y = e^{- 8 x}, x \geq 0

What is the integral of (e^{-x}-e^5)(e^x+e^{-3}) ?
The integral of (e^{-x}-e^5)(e^x+e^{-3}) is x-e^{-x-3}-e^{5+x}-e^2x+C