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

The integral of 3/(sqrt(e^{2a)-4)} is 3/2 arctan(1/2 sqrt(e^{2a)-4})+C

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

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

\frac{3}{2} arctan (\frac{1}{2} e^{2 a} - 4) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Apply Integral Substitution

= 3 \cdot \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 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 3 \cdot \frac{1}{2} arctan (v)

Substitute back

= 3 \cdot \frac{1}{2} arctan (\frac{e ^{2 a} - 4}{2})

Simplify

3 \cdot \frac{1}{2} arctan (\frac{e ^{2 a} - 4}{2}) : \frac{3}{2} arctan (\frac{1}{2} e^{2 a} - 4)

= \frac{3}{2} arctan (\frac{1}{2} e^{2 a} - 4)

Add a constant to the solution

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

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