What is the integral of (e^x)/(e^{2x)-9} ?

The integral of (e^x)/(e^{2x)-9} is -1/6 (ln|(e^x)/3+1|-ln|(e^x)/3-1|)+C

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integral of (e^x)/(e^{2x)-9}

\int \frac{e ^{x}}{e ^{2 x} - 9} d x

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

Solution steps

\int \frac{e ^{x}}{e ^{2 x} - 9} d x

Factor

e^{2 x} - 9 : - (- e^{2 x} + 9)

= \int \frac{e ^{x}}{- ( - e ^{2 x} + 9 )} d x

Take the constant out:

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

= - \int \frac{e ^{x}}{- e ^{2 x} + 9} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 9} d u

= - \int \frac{1}{- u ^{2} + 9} d u

Apply Integral Substitution

= - \int \frac{1}{3 ( - v ^{2} + 1 )} d v

Take the constant out:

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

= - \frac{1}{3} \cdot \int \frac{1}{- v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= - \frac{1}{3} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= - \frac{1}{3} (\frac{ln \frac{e ^{x}}{3} + 1}{2} - \frac{ln \frac{e ^{x}}{3} - 1}{2})

Simplify

- \frac{1}{3} (\frac{ln \frac{e ^{x}}{3} + 1}{2} - \frac{ln \frac{e ^{x}}{3} - 1}{2}) : - \frac{1}{6} (ln \frac{e ^{x}}{3} + 1 - ln \frac{e ^{x}}{3} - 1)

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

Add a constant to the solution

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

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What is the integral of (e^x)/(e^{2x)-9} ?
The integral of (e^x)/(e^{2x)-9} is -1/6 (ln|(e^x)/3+1|-ln|(e^x)/3-1|)+C