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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integral Substitution

= \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

= \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}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

n \to \infty lim (3^{- n})

y = (x^{3} - 2)^{2}

(x)^{(\frac{1}{3})}

\frac{\partial}{\partial x} (x^{7} y z^{2} + 3 yz)

\frac{d}{d x} (4 sin (5 x))

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