What is the integral of 1/(e^x-9e^{-x)} ?

The integral of 1/(e^x-9e^{-x)} is -1/6 ln|e^x+3|+1/6 ln|e^x-3|+C

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integral of 1/(e^x-9e^{-x)}

\int \frac{1}{e ^{x} - 9 e ^{- x}} d x

- \frac{1}{6} ln ∣ e^{x} + 3 ∣ + \frac{1}{6} ln ∣ e^{x} - 3 ∣ + C

Solution steps

\int \frac{1}{e ^{x} - 9 e ^{- x}} d x

Factor

e^{x} - 9 e^{- x} : e^{- x} (e^{x} + 3) (e^{x} - 3)

= \int \frac{1}{e ^{- x} ( e ^{x} + 3 ) ( e ^{x} - 3 )} d x

Apply exponent rule:

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

\frac{1}{e ^{- x} ( e ^{x} + 3 ) ( e ^{x} - 3 )} = e^{x} \frac{1}{( e ^{x} + 3 ) ( e ^{x} - 3 )}

= \int e^{x} \frac{1}{( e ^{x} + 3 ) ( e ^{x} - 3 )} d x

Simplify

e^{x} \frac{1}{( e ^{x} + 3 ) ( e ^{x} - 3 )} : \frac{e ^{x}}{( e ^{x} + 3 ) ( e ^{x} - 3 )}

= \int \frac{e ^{x}}{( e ^{x} + 3 ) ( e ^{x} - 3 )} d x

Apply u-substitution

= \int \frac{1}{( u + 3 ) ( u - 3 )} d u

Take the partial fraction of

\frac{1}{( u + 3 ) ( u - 3 )} : - \frac{1}{6 ( u + 3 )} + \frac{1}{6 ( u - 3 )}

= \int - \frac{1}{6 ( u + 3 )} + \frac{1}{6 ( u - 3 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{6 ( u + 3 )} d u + \int \frac{1}{6 ( u - 3 )} d u

\int \frac{1}{6 ( u + 3 )} d u = \frac{1}{6} ln ∣ u + 3 ∣

\int \frac{1}{6 ( u - 3 )} d u = \frac{1}{6} ln ∣ u - 3 ∣

= - \frac{1}{6} ln ∣ u + 3 ∣ + \frac{1}{6} ln ∣ u - 3 ∣

Substitute back

u = e^{x}

= - \frac{1}{6} ln ∣ e^{x} + 3 ∣ + \frac{1}{6} ln ∣ e^{x} - 3 ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ e^{x} + 3 ∣ + \frac{1}{6} ln ∣ e^{x} - 3 ∣ + C

What is the integral of 1/(e^x-9e^{-x)} ?
The integral of 1/(e^x-9e^{-x)} is -1/6 ln|e^x+3|+1/6 ln|e^x-3|+C