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 2.12764…*1.6^x-7.15502…*1.15^x+C

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integral of (3.2^x-2.3^x)/(2^x)

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

2.12764 \dots \cdot 1. 6^{x} - 7.15502 \dots \cdot 1.1 5^{x} + C

Solution steps

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

Expand

\frac{3. 2 ^{x} - 2. 3 ^{x}}{2 ^{x}} : (\frac{3.2}{2})^{x} - (\frac{2.3}{2})^{x}

= \int (\frac{3.2}{2})^{x} - (\frac{2.3}{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{3.2}{2})^{x} d x - \int (\frac{2.3}{2})^{x} d x

\int (\frac{3.2}{2})^{x} d x = 2.12764 \dots \cdot 1. 6^{x}

\int (\frac{2.3}{2})^{x} d x = 7.15502 \dots \cdot 1.1 5^{x}

= 2.12764 \dots \cdot 1. 6^{x} - 7.15502 \dots \cdot 1.1 5^{x}

Add a constant to the solution

= 2.12764 \dots \cdot 1. 6^{x} - 7.15502 \dots \cdot 1.1 5^{x} + C

\int \frac{42 x}{3 x ^{2} + e} d x

y^{^{'}} + y + 2 sin (x) = 0

e^{2 x} + 2 x

\frac{d}{d x} (\frac{1}{3 8 x ^{2}})

\int_{1}^{e} ln (8 x) 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 2.12764…*1.6^x-7.15502…*1.15^x+C