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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= - (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = 1 - e^{2 x}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (2 lo g_{6} (x))

y^{^{''}} + 3 y^{^{'}} + 2.25 y = 0

\frac{d}{d x} (5 e^{x} + \frac{4}{3 x})

\frac{d y}{d t} = 7 y^{2} t^{3}, y (2) = 3

\int_{1}^{3} \frac{t}{3 + t ^{2}} d t

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