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

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

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integral of (2^{x+1}-3^{x+1})/(2^{x+1)}

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

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

Solution steps

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

Apply u-substitution

: \int \frac{2 ^{u} - 3 ^{u}}{2 ^{u}} d u

= \int \frac{2 ^{u} - 3 ^{u}}{2 ^{u}} d u

Expand

\frac{2 ^{u} - 3 ^{u}}{2 ^{u}} : 1 - (\frac{3}{2})^{u}

= \int 1 - (\frac{3}{2})^{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d u - \int (\frac{3}{2})^{u} d u

\int 1 d u = u

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

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

Substitute back

u = x + 1

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

Add a constant to the solution

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

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