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

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

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integral of (2^x+e^x)

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

\frac{2 ^{x}}{ln ( 2 )} + e^{x} + C

Solution steps

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

\int 2^{x} d x = \frac{2 ^{x}}{ln ( 2 )}

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

= \frac{2 ^{x}}{ln ( 2 )} + e^{x}

Add a constant to the solution

= \frac{2 ^{x}}{ln ( 2 )} + e^{x} + C

\int_{- 3}^{3} \frac{14}{x ^{2} - 6 x - 40} d x

\frac{\partial}{\partial x} (\frac{( y ^{3} )}{x})

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

\int_{0}^{1} \frac{22}{4 y - 1} d y

\frac{d}{d x} (\frac{x ^{2} - 1}{x ^{2} - 4})

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