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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Factor

u^{2} - 1 : - (- u^{2} + 1)

= \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 = e^{2 x} + 1

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (x), y (0) = 1, y^{^{'}} (0) = - 1

y^{^{'}} = x + y, y (0) = 7

\frac{d}{d x} (\frac{4 x ^{2} + 2 x + 8}{x})

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

y^{^{''}} - 8 y^{^{'}} + 20 y = 0

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