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

The integral of (2^{3x-2})/(3^{2x-3)} is (8^x*3^{3-2x})/(4ln(8/9))+C

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

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

\frac{8 ^{x} \cdot 3 ^{3 - 2 x}}{4 ln ( \frac{8}{9} )} + C

Solution steps

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

\frac{2 ^{3 x - 2}}{3 ^{2 x - 3}} = \frac{2 ^{- 2}}{3 ^{- 3}} \frac{2 ^{3 x}}{3 ^{2 x}}

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

\frac{2 ^{- 2}}{3 ^{- 3}} = \frac{27}{4}

= \frac{27}{4} \cdot \int \frac{2 ^{3 x}}{3 ^{2 x}} d x

Simplify

\frac{2 ^{3 x}}{3 ^{2 x}} : (\frac{8}{9})^{x}

= \frac{27}{4} \cdot \int (\frac{8}{9})^{x} d x

\int a^{x} d x = \frac{a ^{x}}{ln a}

= \frac{27}{4} \cdot \frac{( \frac{8}{9} ) ^{x}}{ln ( \frac{8}{9} )}

Simplify

\frac{27}{4} \cdot \frac{( \frac{8}{9} ) ^{x}}{ln ( \frac{8}{9} )} : \frac{8 ^{x} \cdot 3 ^{3 - 2 x}}{4 ln ( \frac{8}{9} )}

= \frac{8 ^{x} \cdot 3 ^{3 - 2 x}}{4 ln ( \frac{8}{9} )}

Add a constant to the solution

= \frac{8 ^{x} \cdot 3 ^{3 - 2 x}}{4 ln ( \frac{8}{9} )} + C

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What is the integral of (2^{3x-2})/(3^{2x-3)} ?
The integral of (2^{3x-2})/(3^{2x-3)} is (8^x*3^{3-2x})/(4ln(8/9))+C