What is the integral of 7^{x-1} ?

The integral of 7^{x-1} is (7^{x-1})/(ln(7))+C

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integral of 7^{x-1}

\int 7^{x - 1} d x

\frac{7 ^{x - 1}}{ln ( 7 )} + C

Solution steps

\int 7^{x - 1} d x

Simplify

7^{x - 1} : 7^{x} \cdot 7^{- 1}

= \int 7^{x} \cdot 7^{- 1} d x

Take the constant out:

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

= 7^{- 1} \cdot \int 7^{x} d x

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{7} \cdot \int 7^{x} d x

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

= \frac{1}{7} \cdot \frac{7 ^{x}}{ln ( 7 )}

Simplify

\frac{1}{7} \cdot \frac{7 ^{x}}{ln ( 7 )} : \frac{7 ^{x - 1}}{ln ( 7 )}

= \frac{7 ^{x - 1}}{ln ( 7 )}

Add a constant to the solution

= \frac{7 ^{x - 1}}{ln ( 7 )} + C

What is the integral of 7^{x-1} ?
The integral of 7^{x-1} is (7^{x-1})/(ln(7))+C