What is the integral of (e^x)/(36-e^{2x)} ?

The integral of (e^x)/(36-e^{2x)} is 1/12 (ln|(e^x)/6+1|-ln|(e^x)/6-1|)+C

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integral of (e^x)/(36-e^{2x)}

\int \frac{e ^{x}}{36 - e ^{2 x}} d x

\frac{1}{12} (ln \frac{e ^{x}}{6} + 1 - ln \frac{e ^{x}}{6} - 1) + C

Solution steps

\int \frac{e ^{x}}{36 - e ^{2 x}} d x

Apply u-substitution

: \int \frac{1}{36 - u ^{2}} d u

= \int \frac{1}{36 - u ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{6 ( - 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}{6} \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}{6} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

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

Simplify

\frac{1}{6} (\frac{ln \frac{e ^{x}}{6} + 1}{2} - \frac{ln \frac{e ^{x}}{6} - 1}{2}) : \frac{1}{12} (ln \frac{e ^{x}}{6} + 1 - ln \frac{e ^{x}}{6} - 1)

= \frac{1}{12} (ln \frac{e ^{x}}{6} + 1 - ln \frac{e ^{x}}{6} - 1)

Add a constant to the solution

= \frac{1}{12} (ln \frac{e ^{x}}{6} + 1 - ln \frac{e ^{x}}{6} - 1) + C

\int x 2 - x d x

\frac{d}{d x} (\frac{2 x - 1}{x ^{2} + 1})

x \to 2 lim (\frac{0}{x - 2})

n = 1 \sum \infty n^{3} \cdot 2^{- n}

What is the integral of (e^x)/(36-e^{2x)} ?
The integral of (e^x)/(36-e^{2x)} is 1/12 (ln|(e^x)/6+1|-ln|(e^x)/6-1|)+C