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

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

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integral of ((a^x))/((a^x-1))

integral

\frac{( a ^{x} )}{( a ^{x} - 1 )}

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ln ( a )} d u

= \int \frac{1}{u ln ( a )} d u

Take the constant out:

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

= \frac{1}{ln ( a )} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{ln ( a )} ln ∣ u ∣

Substitute back

u = a^{x} - 1

= \frac{1}{ln ( a )} ln ∣ a^{x} - 1 ∣

Add a constant to the solution

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

sec^{4} (x) tan^{2} (x)

e^{x} (\frac{1}{x} - \frac{1}{x ^{2}})

3 x^{2} + 6 x - 2

(2^{x})^{2}

y = 2 x

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