What is the integral of a^{x+1} ?

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

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integral of a^{x+1}

\int a^{x + 1} d x

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

Solution steps

\int a^{x + 1} d x

Simplify

a^{x + 1} : a^{x} a

= \int a^{x} a d x

Take the constant out:

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

= a \cdot \int a^{x} d x

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

= a \frac{a ^{x}}{ln ( a )}

Simplify

a \frac{a ^{x}}{ln ( a )} : \frac{a ^{x + 1}}{ln ( a )}

= \frac{a ^{x + 1}}{ln ( a )}

Add a constant to the solution

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

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