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

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

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

\int \frac{1}{e ^{2 t} - 2} d t

\frac{1}{2} arctan (\frac{1}{2} (e^{2 t} - 2)) + C

Solution steps

\int \frac{1}{e ^{2 t} - 2} d t

Apply u-substitution

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

= \int \frac{1}{2 e ^{u} - 2} 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{1}{e ^{u} - 2} d u

Apply u-substitution

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

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

Apply Integral Substitution

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

Use the common integral:

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

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

Substitute back

= \frac{1}{2} \cdot 2 \cdot \frac{1}{2} arctan (\frac{e ^{2 t} - 2}{2})

Simplify

\frac{1}{2} \cdot 2 \cdot \frac{1}{2} arctan (\frac{e ^{2 t} - 2}{2}) : \frac{1}{2} arctan (\frac{1}{2} (e^{2 t} - 2))

= \frac{1}{2} arctan (\frac{1}{2} (e^{2 t} - 2))

Add a constant to the solution

= \frac{1}{2} arctan (\frac{1}{2} (e^{2 t} - 2)) + C

\frac{d}{d x} (0.6 x^{3} 4^{x})

y = 3 x^{2}, y = 2, x = 0, x \geq 0

\frac{d}{d x} (x e^{- \frac{x ^{2}}{2}})

\frac{x}{2 + x ^{2}}

(3 x + 7) x^{2}

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