What is the integral of 1/(a^x) ?

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

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integral of 1/(a^x)

integral

\frac{1}{a ^{x}}

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

Solution steps

\int \frac{1}{a ^{x}} d x

Apply exponent rule:

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

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

= \int a^{- x} d x

Apply u-substitution

: \int - a^{u} d u

= \int - a^{u} d u

Take the constant out:

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

= - \int a^{u} d u

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

= - \frac{a ^{u}}{ln ( a )}

Substitute back

u = - x

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

Add a constant to the solution

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

\frac{1}{( x ^{2} + 2 )}

\frac{x ^{2}}{( x ^{3} + 1 )}

\frac{x}{( x + 2 )}

\frac{( 2 x )}{( 1 + 4 x ^{2} )}

\frac{1}{( x ^{3} + x )}

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