What is the integral of (e^x)/(25-e^{2x)} ?

The integral of (e^x)/(25-e^{2x)} is 1/10 (ln|(e^x)/5+1|-ln|(e^x)/5-1|)+C

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integral of (e^x)/(25-e^{2x)}

\int \frac{e ^{x}}{25 - e ^{2 x}} d x

\frac{1}{10} (ln \frac{e ^{x}}{5} + 1 - ln \frac{e ^{x}}{5} - 1) + C

Solution steps

\int \frac{e ^{x}}{25 - e ^{2 x}} d x

Apply u-substitution

: \int \frac{1}{25 - u ^{2}} d u

= \int \frac{1}{25 - u ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{5 ( - 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}{5} \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}{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}{10} (ln \frac{e ^{x}}{5} + 1 - ln \frac{e ^{x}}{5} - 1)

= \frac{1}{10} (ln \frac{e ^{x}}{5} + 1 - ln \frac{e ^{x}}{5} - 1)

Add a constant to the solution

= \frac{1}{10} (ln \frac{e ^{x}}{5} + 1 - ln \frac{e ^{x}}{5} - 1) + C

x \to + (- 1) lim (\frac{k x + 3}{4})

\frac{d}{d x} (a x^{2} e^{5 x})

t^{4} + 3 e^{3 t} t^{2} + e^{- t} t + 1

- \frac{4 x}{x ^{2} + 1}

x \to \infty lim (x + 8)

What is the integral of (e^x)/(25-e^{2x)} ?
The integral of (e^x)/(25-e^{2x)} is 1/10 (ln|(e^x)/5+1|-ln|(e^x)/5-1|)+C