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

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

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integral of (e^{-x}-e^5)(e^x+e^{-2})

\int (e^{- x} - e^{5}) (e^{x} + e^{- 2}) d x

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

Solution steps

\int (e^{- x} - e^{5}) (e^{x} + e^{- 2}) d x

Expand

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

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

\int 1 d x = x

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

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

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

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

Add a constant to the solution

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

\int x 4^{3 x^{2}} d x

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

\int \frac{x + 1}{x ^{2} - 2 x + 1} d x

\frac{( x ^{2} - 1 )}{x ^{2} + x + 1}

2 x y y^{^{'}} = x^{2} + 2 y^{2}

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