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

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

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integral of ((32^x-23^x))/(2^x)

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

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

Solution steps

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

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

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

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

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

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

Add a constant to the solution

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

y = 3 (x^{2} - x + 1) (5 x - 3)

\frac{\partial}{\partial x} ((x + y^{2})^{3})

\frac{4}{x}, at x = - 5

\int 5 x cos (3 x) d x

\int \frac{1}{( 4 x ^{2} - 4 x - 3 ) ^{\frac{3}{2}}} d x

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