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

The integral of (e^{8x})/(e^{8x)-9e^{-8x}} is 1/16 ln|e^{16x}-9|+C

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

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

\frac{1}{16} ln e^{16 x} - 9 + C

Solution steps

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

Apply u-substitution

: \int \frac{e ^{2 u}}{8 ( e ^{2 u} - 9 )} d u

= \int \frac{e ^{2 u}}{8 ( e ^{2 u} - 9 )} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{e ^{2 u}}{e ^{2 u} - 9} d u

Apply u-substitution

: \int \frac{1}{2 v} d v

= \frac{1}{8} \cdot \int \frac{1}{2 v} d v

Take the constant out:

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

= \frac{1}{8} \cdot \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{8} \cdot \frac{1}{2} ln ∣ v ∣

Substitute back

= \frac{1}{8} \cdot \frac{1}{2} ln e^{2 \cdot 8 x} - 9

Simplify

\frac{1}{8} \cdot \frac{1}{2} ln e^{2 \cdot 8 x} - 9 : \frac{1}{16} ln e^{16 x} - 9

= \frac{1}{16} ln e^{16 x} - 9

Add a constant to the solution

= \frac{1}{16} ln e^{16 x} - 9 + C

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What is the integral of (e^{8x})/(e^{8x)-9e^{-8x}} ?
The integral of (e^{8x})/(e^{8x)-9e^{-8x}} is 1/16 ln|e^{16x}-9|+C