What is the integral of e^x(2^x-1) ?

The integral of e^x(2^x-1) is ((2e)^x)/(ln(2e))-e^x+C

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integral of e^x(2^x-1)

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

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

Solution steps

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

Expand

e^{x} (2^{x} - 1) : 2^{x} e^{x} - e^{x}

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

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

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

Add a constant to the solution

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

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What is the integral of e^x(2^x-1) ?
The integral of e^x(2^x-1) is ((2e)^x)/(ln(2e))-e^x+C