What is the integral of (3^{x+2})/(4^{x-1)} ?

The integral of (3^{x+2})/(4^{x-1)} is (2^{2-2x}*3^{x+2})/(ln(3/4))+C

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integral of (3^{x+2})/(4^{x-1)}

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

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

Solution steps

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

\frac{3 ^{x + 2}}{4 ^{x - 1}} = \frac{3 ^{2}}{4 ^{- 1}} \frac{3 ^{x}}{4 ^{x}}

= \int \frac{3 ^{2}}{4 ^{- 1}} \cdot \frac{3 ^{x}}{4 ^{x}} d x

Take the constant out:

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

= \frac{3 ^{2}}{4 ^{- 1}} \cdot \int \frac{3 ^{x}}{4 ^{x}} d x

\frac{3 ^{2}}{4 ^{- 1}} = 36

= 36 \cdot \int \frac{3 ^{x}}{4 ^{x}} d x

Apply exponent rule:

\frac{a ^{c}}{b ^{c}} = (\frac{a}{b})^{c}

= 36 \cdot \int (\frac{3}{4})^{x} d x

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

= 36 \cdot \frac{( \frac{3}{4} ) ^{x}}{ln ( \frac{3}{4} )}

Simplify

36 \cdot \frac{( \frac{3}{4} ) ^{x}}{ln ( \frac{3}{4} )} : \frac{2 ^{2 - 2 x} \cdot 3 ^{x + 2}}{ln ( \frac{3}{4} )}

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

Add a constant to the solution

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

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What is the integral of (3^{x+2})/(4^{x-1)} ?
The integral of (3^{x+2})/(4^{x-1)} is (2^{2-2x}*3^{x+2})/(ln(3/4))+C