What is the integral of (8e^x)/(e^{2x)-16} ?

The integral of (8e^x)/(e^{2x)-16} is -ln| 1/4 e^x+1|+ln| 1/4 e^x-1|+C

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integral of (8e^x)/(e^{2x)-16}

\int \frac{8 e ^{x}}{e ^{2 x} - 16} d x

- ln \frac{1}{4} e^{x} + 1 + ln \frac{1}{4} e^{x} - 1 + C

Solution steps

\int \frac{8 e ^{x}}{e ^{2 x} - 16} d x

Take the constant out:

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

= 8 \cdot \int \frac{e ^{x}}{e ^{2 x} - 16} d x

Factor

e^{2 x} - 16 : - (- e^{2 x} + 16)

= 8 \cdot \int \frac{e ^{x}}{- ( - e ^{2 x} + 16 )} d x

Take the constant out:

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

= 8 (- \int \frac{e ^{x}}{- e ^{2 x} + 16} d x)

Apply u-substitution

: \int \frac{1}{- u ^{2} + 16} d u

= 8 (- \int \frac{1}{- u ^{2} + 16} d u)

Apply Integral Substitution

= 8 (- \int \frac{1}{4 ( - v ^{2} + 1 )} d v)

Take the constant out:

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

= 8 (- \frac{1}{4} \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}

= 8 (- \frac{1}{4} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 8 (- \frac{1}{4} (\frac{ln \frac{e ^{x}}{4} + 1}{2} - \frac{ln \frac{e ^{x}}{4} - 1}{2}))

Simplify

8 (- \frac{1}{4} (\frac{ln \frac{e ^{x}}{4} + 1}{2} - \frac{ln \frac{e ^{x}}{4} - 1}{2})) : - ln \frac{1}{4} e^{x} + 1 + ln \frac{1}{4} e^{x} - 1

= - ln \frac{1}{4} e^{x} + 1 + ln \frac{1}{4} e^{x} - 1

Add a constant to the solution

= - ln \frac{1}{4} e^{x} + 1 + ln \frac{1}{4} e^{x} - 1 + C

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What is the integral of (8e^x)/(e^{2x)-16} ?
The integral of (8e^x)/(e^{2x)-16} is -ln| 1/4 e^x+1|+ln| 1/4 e^x-1|+C