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

The integral of (e^x)/(sqrt(e^{2x)-2)} is ln| 1/(sqrt(2))sqrt(e^{2x)-2}+sqrt(1/2 e^{2x)}|+C

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

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

ln \frac{1}{2} e^{2 x} - 2 + \frac{1}{2} e^{2 x} + C

Solution steps

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln tan (arctan (\frac{1}{2} e^{2 x} - 2)) + sec (arctan (\frac{1}{2} e^{2 x} - 2))

Simplify

ln tan (arctan (\frac{1}{2} e^{2 x} - 2)) + sec (arctan (\frac{1}{2} e^{2 x} - 2)) : ln \frac{1}{2} e^{2 x} - 2 + \frac{1}{2} e^{2 x}

= ln \frac{1}{2} e^{2 x} - 2 + \frac{1}{2} e^{2 x}

Add a constant to the solution

= ln \frac{1}{2} e^{2 x} - 2 + \frac{1}{2} e^{2 x} + C

x \to π lim ((π - x) csc (x))

9 x y^{2} - 3 + (9 x^{2} y + 4) \frac{d y}{d x} = 0

\frac{d y}{d x} + 4 y = 6

\int (x - 2)^{- 3} d x

\frac{\partial}{\partial y} (2)

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