What is the integral of 2xe^x-e^x-e^{-x} ?

The integral of 2xe^x-e^x-e^{-x} is 2e^xx-3e^x+e^{-x}+C

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integral of 2xe^x-e^x-e^{-x}

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

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

Solution steps

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int 5 x^{6} (2 x^{7} + 5)^{4} d x

x^{3} + 3 x^{2} + 2 x, - 3, 3

e^{x}, 2 x - 1, x = - 1, x = 2

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

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

What is the integral of 2xe^x-e^x-e^{-x} ?
The integral of 2xe^x-e^x-e^{-x} is 2e^xx-3e^x+e^{-x}+C