What is the integral of (6^x-2^x)/(3^x) ?

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

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integral of (6^x-2^x)/(3^x)

\int \frac{6 ^{x} - 2 ^{x}}{3 ^{x}} d x

\frac{2 ^{x}}{ln ( 2 )} - \frac{2 ^{x}}{ln ( \frac{2}{3} ) \cdot 3 ^{x}} + C

Solution steps

\int \frac{6 ^{x} - 2 ^{x}}{3 ^{x}} d x

Expand

\frac{6 ^{x} - 2 ^{x}}{3 ^{x}} : 2^{x} - (\frac{2}{3})^{x}

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

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

\int (\frac{2}{3})^{x} d x = \frac{2 ^{x}}{ln ( \frac{2}{3} ) \cdot 3 ^{x}}

= \frac{2 ^{x}}{ln ( 2 )} - \frac{2 ^{x}}{ln ( \frac{2}{3} ) \cdot 3 ^{x}}

Add a constant to the solution

= \frac{2 ^{x}}{ln ( 2 )} - \frac{2 ^{x}}{ln ( \frac{2}{3} ) \cdot 3 ^{x}} + C

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