What is the integral of (e^{9x})/(e^{9x)+6} ?

The integral of (e^{9x})/(e^{9x)+6} is 1/9 ln|e^{9x}+6|+C

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integral of (e^{9x})/(e^{9x)+6}

\int \frac{e ^{9 x}}{e ^{9 x} + 6} d x

\frac{1}{9} ln e^{9 x} + 6 + C

Solution steps

\int \frac{e ^{9 x}}{e ^{9 x} + 6} d x

Apply u-substitution

= \int \frac{1}{9 u} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{9} ln ∣ u ∣

Substitute back

u = e^{9 x} + 6

= \frac{1}{9} ln e^{9 x} + 6

Add a constant to the solution

= \frac{1}{9} ln e^{9 x} + 6 + C

What is the integral of (e^{9x})/(e^{9x)+6} ?
The integral of (e^{9x})/(e^{9x)+6} is 1/9 ln|e^{9x}+6|+C