What is the integral of 3^x(1-2^x) ?

The integral of 3^x(1-2^x) is (3^x)/(ln(3))-(6^x)/(ln(6))+C

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integral of 3^x(1-2^x)

\int 3^{x} (1 - 2^{x}) d x

\frac{3 ^{x}}{ln ( 3 )} - \frac{6 ^{x}}{ln ( 6 )} + C

Solution steps

\int 3^{x} (1 - 2^{x}) d x

Expand

3^{x} (1 - 2^{x}) : 3^{x} - 6^{x}

= \int 3^{x} - 6^{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 3^{x} d x - \int 6^{x} d x

\int 3^{x} d x = \frac{3 ^{x}}{ln ( 3 )}

\int 6^{x} d x = \frac{6 ^{x}}{ln ( 6 )}

= \frac{3 ^{x}}{ln ( 3 )} - \frac{6 ^{x}}{ln ( 6 )}

Add a constant to the solution

= \frac{3 ^{x}}{ln ( 3 )} - \frac{6 ^{x}}{ln ( 6 )} + C

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What is the integral of 3^x(1-2^x) ?
The integral of 3^x(1-2^x) is (3^x)/(ln(3))-(6^x)/(ln(6))+C