What is the integral of (e^{9x})/(e^{9x)+8} ?

The integral of (e^{9x})/(e^{9x)+8} is 1/9 ln|e^{9x}+8|+C

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integral of (e^{9x})/(e^{9x)+8}

\int \frac{e ^{9 x}}{e ^{9 x} + 8} d x

\frac{1}{9} ln e^{9 x} + 8 + C

Solution steps

\int \frac{e ^{9 x}}{e ^{9 x} + 8} d x

Apply u-substitution

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

= \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} + 8

= \frac{1}{9} ln e^{9 x} + 8

Add a constant to the solution

= \frac{1}{9} ln e^{9 x} + 8 + C

y = e^{x}, y = x e^{x}, x = 0

(\frac{- ln ( 1 + 6 x )}{x})^{^{'}}

\int (2 x - 1) e^{x} d x

f (x) = x^{2} - 2 x + 3

f (x) = 9 x - 6 x, at x = 1

What is the integral of (e^{9x})/(e^{9x)+8} ?
The integral of (e^{9x})/(e^{9x)+8} is 1/9 ln|e^{9x}+8|+C