What is the integral of 3/(sqrt(4-e^{2x))} ?

The integral of 3/(sqrt(4-e^{2x))} is -3/4 (ln|(sqrt(4-e^{2x)})/2+1|-ln|(sqrt(4-e^{2x)})/2-1|)+C

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integral of 3/(sqrt(4-e^{2x))}

\int \frac{3}{4 - e ^{2 x}} d x

- \frac{3}{4} (ln \frac{4 - e ^{2 x}}{2} + 1 - ln \frac{4 - e ^{2 x}}{2} - 1) + C

Solution steps

\int \frac{3}{4 - e ^{2 x}} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{4 - e ^{2 x}} d x

Apply u-substitution

: \int - \frac{1}{- u ^{2} + 4} d u

= 3 \cdot \int - \frac{1}{- u ^{2} + 4} d u

Take the constant out:

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

= 3 (- \int \frac{1}{- u ^{2} + 4} d u)

Apply Integral Substitution

= 3 (- \int \frac{1}{2 ( - v ^{2} + 1 )} d v)

Take the constant out:

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

= 3 (- \frac{1}{2} \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}

= 3 (- \frac{1}{2} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 3 - \frac{1}{2} \frac{ln \frac{4 - e ^{2 x}}{2} + 1}{2} - \frac{ln \frac{4 - e ^{2 x}}{2} - 1}{2}

Simplify

3 - \frac{1}{2} \frac{ln \frac{4 - e ^{2 x}}{2} + 1}{2} - \frac{ln \frac{4 - e ^{2 x}}{2} - 1}{2} : - \frac{3}{4} (ln \frac{4 - e ^{2 x}}{2} + 1 - ln \frac{4 - e ^{2 x}}{2} - 1)

= - \frac{3}{4} (ln \frac{4 - e ^{2 x}}{2} + 1 - ln \frac{4 - e ^{2 x}}{2} - 1)

Add a constant to the solution

= - \frac{3}{4} (ln \frac{4 - e ^{2 x}}{2} + 1 - ln \frac{4 - e ^{2 x}}{2} - 1) + C

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What is the integral of 3/(sqrt(4-e^{2x))} ?
The integral of 3/(sqrt(4-e^{2x))} is -3/4 (ln|(sqrt(4-e^{2x)})/2+1|-ln|(sqrt(4-e^{2x)})/2-1|)+C