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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u}{2 ( u + 9 )} d u

= \int \frac{u}{2 ( u + 9 )} d u

Take the constant out:

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

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

Apply u-substitution

: \int \frac{2 v ^{2}}{v ^{2} + 9} d v

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

Take the constant out:

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

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

\frac{v ^{2}}{v ^{2} + 9} = - \frac{9}{v ^{2} + 9} + 1

= \frac{1}{2} \cdot 2 \cdot \int - \frac{9}{v ^{2} + 9} + 1 d v

Apply Integral Substitution

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

Take the constant out:

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d w = w

= \frac{1}{2} \cdot 2 \cdot 3 (- arctan (w) + w)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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