What is the integral of (3*2^x+4*3^x)/(2^x) ?

The integral of (3*2^x+4*3^x)/(2^x) is 3x+(3^x*2^{2-x})/(ln(3/2))+C

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integral of (32^x+43^x)/(2^x)

\int \frac{3 \cdot 2 ^{x} + 4 \cdot 3 ^{x}}{2 ^{x}} d x

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

Solution steps

\int \frac{3 \cdot 2 ^{x} + 4 \cdot 3 ^{x}}{2 ^{x}} d x

\frac{3 \cdot 2 ^{x} + 4 \cdot 3 ^{x}}{2 ^{x}} = (\frac{3 \cdot 2 ^{x}}{2 ^{x}} + \frac{4 \cdot 3 ^{x}}{2 ^{x}})

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

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

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

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

Add a constant to the solution

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

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What is the integral of (3*2^x+4*3^x)/(2^x) ?
The integral of (3*2^x+4*3^x)/(2^x) is 3x+(3^x*2^{2-x})/(ln(3/2))+C