What is the integral of 1/(sqrt(e^{2x)-9)} ?

The integral of 1/(sqrt(e^{2x)-9)} is 1/3 arctan((sqrt(e^{2x)-9})/3)+C

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integral of 1/(sqrt(e^{2x)-9)}

\int \frac{1}{e ^{2 x} - 9} d x

\frac{1}{3} arctan (\frac{e ^{2 x} - 9}{3}) + C

Solution steps

\int \frac{1}{e ^{2 x} - 9} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 9} d u

= \int \frac{1}{u ^{2} + 9} d u

Apply Integral Substitution

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

Use the common integral:

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

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

Substitute back

= \frac{1}{3} arctan (\frac{e ^{2 x} - 9}{3})

Add a constant to the solution

= \frac{1}{3} arctan (\frac{e ^{2 x} - 9}{3}) + C

\int \frac{1}{x} \frac{x + 1}{x} d x

\int \frac{5 x - 4}{x ^{3} + 4 x} d x

\int (x^{2} + 1)^{3} 2 x d x

What is the integral of 1/(sqrt(e^{2x)-9)} ?
The integral of 1/(sqrt(e^{2x)-9)} is 1/3 arctan((sqrt(e^{2x)-9})/3)+C