What is the integral of (e^x)/(5-e^{2x)} ?

The integral of (e^x)/(5-e^{2x)} is 1/(2sqrt(5))(ln|(e^x)/(sqrt(5))+1|-ln|(e^x)/(sqrt(5))-1|)+C

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integral of (e^x)/(5-e^{2x)}

\int \frac{e ^{x}}{5 - e ^{2 x}} d x

\frac{1}{2 5} (ln \frac{e ^{x}}{5} + 1 - ln \frac{e ^{x}}{5} - 1) + C

Solution steps

\int \frac{e ^{x}}{5 - e ^{2 x}} d x

Apply u-substitution

: \int \frac{1}{5 - u ^{2}} d u

= \int \frac{1}{5 - u ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{5 ( 1 - v ^{2} )} d v

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{1}{1 - v ^{2}} d v

Use the common integral:

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

= \frac{1}{5} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= \frac{1}{5} \frac{ln \frac{e ^{x}}{5} + 1}{2} - \frac{ln \frac{e ^{x}}{5} - 1}{2}

Simplify

\frac{1}{5} \frac{ln \frac{e ^{x}}{5} + 1}{2} - \frac{ln \frac{e ^{x}}{5} - 1}{2} : \frac{1}{2 5} (ln \frac{e ^{x}}{5} + 1 - ln \frac{e ^{x}}{5} - 1)

= \frac{1}{2 5} (ln \frac{e ^{x}}{5} + 1 - ln \frac{e ^{x}}{5} - 1)

Add a constant to the solution

= \frac{1}{2 5} (ln \frac{e ^{x}}{5} + 1 - ln \frac{e ^{x}}{5} - 1) + C

x \to \infty lim (- \frac{3}{2} x)

x \to - 2 + lim (\frac{x ^{2} - 9}{x ^{2} - 2 x - 8})

n = 1 \sum \infty \frac{9 ^{n}}{n !}

x \to \infty lim (3 x + x^{2})

(0, 3), (3, 0)

What is the integral of (e^x)/(5-e^{2x)} ?
The integral of (e^x)/(5-e^{2x)} is 1/(2sqrt(5))(ln|(e^x)/(sqrt(5))+1|-ln|(e^x)/(sqrt(5))-1|)+C