What is the integral of (e^2-2^e)e^x ?

The integral of (e^2-2^e)e^x is (e^2-2^e)e^x+C

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integral of (e^2-2^e)e^x

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

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

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

Add a constant to the solution

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

z \to 2 i lim (z^{2})

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

\frac{d}{d x} (- x^{- 1})

\int x^{2} cos (x y) d y

f (t) = e^{4 t}

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