What is the integral of sqrt(e^{18x)-4} ?

The integral of sqrt(e^{18x)-4} is -2/9 arctan(1/2 sqrt(e^{18x)-4})+1/9 sqrt(e^{18x)-4}+C

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integral of sqrt(e^{18x)-4}

\int e^{18 x} - 4 d x

- \frac{2}{9} arctan (\frac{1}{2} e^{18 x} - 4) + \frac{1}{9} e^{18 x} - 4 + C

Solution steps

\int e^{18 x} - 4 d x

Apply u-substitution

: \int \frac{u}{18 ( u + 4 )} d u

= \int \frac{u}{18 ( u + 4 )} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int \frac{u}{u + 4} d u

Apply u-substitution

: \int \frac{2 v ^{2}}{v ^{2} + 4} d v

= \frac{1}{18} \cdot \int \frac{2 v ^{2}}{v ^{2} + 4} d v

Take the constant out:

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

= \frac{1}{18} \cdot 2 \cdot \int \frac{v ^{2}}{v ^{2} + 4} d v

\frac{v ^{2}}{v ^{2} + 4} = - \frac{4}{v ^{2} + 4} + 1

= \frac{1}{18} \cdot 2 \cdot \int - \frac{4}{v ^{2} + 4} + 1 d v

Apply Integral Substitution

= \frac{1}{18} \cdot 2 \cdot \int 2 (- \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}{18} \cdot 2 \cdot 2 \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}{18} \cdot 2 \cdot 2 (- \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}{18} \cdot 2 \cdot 2 (- arctan (w) + w)

Substitute back

= \frac{1}{18} \cdot 2 \cdot 2 (- arctan (\frac{e ^{18 x} - 4}{2}) + \frac{e ^{18 x} - 4}{2})

Simplify

\frac{1}{18} \cdot 2 \cdot 2 (- arctan (\frac{e ^{18 x} - 4}{2}) + \frac{e ^{18 x} - 4}{2}) : - \frac{2}{9} arctan (\frac{1}{2} e^{18 x} - 4) + \frac{1}{9} e^{18 x} - 4

= - \frac{2}{9} arctan (\frac{1}{2} e^{18 x} - 4) + \frac{1}{9} e^{18 x} - 4

Add a constant to the solution

= - \frac{2}{9} arctan (\frac{1}{2} e^{18 x} - 4) + \frac{1}{9} e^{18 x} - 4 + C

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What is the integral of sqrt(e^{18x)-4} ?
The integral of sqrt(e^{18x)-4} is -2/9 arctan(1/2 sqrt(e^{18x)-4})+1/9 sqrt(e^{18x)-4}+C