What is the integral of (6^x+1)/(2^x) ?

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

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integral of (6^x+1)/(2^x)

\int \frac{6 ^{x} + 1}{2 ^{x}} d x

\frac{3 ^{x}}{ln ( 3 )} - \frac{1}{ln ( 2 ) \cdot 2 ^{x}} + C

Solution steps

\int \frac{6 ^{x} + 1}{2 ^{x}} d x

Expand

\frac{6 ^{x} + 1}{2 ^{x}} : 3^{x} + \frac{1}{2 ^{x}}

= \int 3^{x} + \frac{1}{2 ^{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 \frac{1}{2 ^{x}} d x

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

\int \frac{1}{2 ^{x}} d x = - \frac{1}{ln ( 2 ) \cdot 2 ^{x}}

= \frac{3 ^{x}}{ln ( 3 )} - \frac{1}{ln ( 2 ) \cdot 2 ^{x}}

Add a constant to the solution

= \frac{3 ^{x}}{ln ( 3 )} - \frac{1}{ln ( 2 ) \cdot 2 ^{x}} + C

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