What is the integral of (3^{x+1}-6^{x-1})/(12^x) ?

The integral of (3^{x+1}-6^{x-1})/(12^x) is -3/(ln(2)*2^{2x+1)}+(2^{-x-1})/(3ln(2))+C

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integral of (3^{x+1}-6^{x-1})/(12^x)

\int \frac{3 ^{x + 1} - 6 ^{x - 1}}{1 2 ^{x}} d x

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

Solution steps

\int \frac{3 ^{x + 1} - 6 ^{x - 1}}{1 2 ^{x}} d x

Expand

\frac{3 ^{x + 1} - 6 ^{x - 1}}{1 2 ^{x}} : \frac{3}{4 ^{x}} - \frac{1}{3} \cdot 2^{- x - 1}

= \int \frac{3}{4 ^{x}} - \frac{1}{3} \cdot 2^{- x - 1} 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 \frac{3}{4 ^{x}} d x - \int \frac{1}{3} \cdot 2^{- x - 1} d x

\int \frac{3}{4 ^{x}} d x = - \frac{3}{ln ( 2 ) \cdot 2 ^{2 x + 1}}

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

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

Simplify

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

Add a constant to the solution

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

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What is the integral of (3^{x+1}-6^{x-1})/(12^x) ?
The integral of (3^{x+1}-6^{x-1})/(12^x) is -3/(ln(2)*2^{2x+1)}+(2^{-x-1})/(3ln(2))+C