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))}

integral

\frac{1}{1 + e ^{- 2 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}{2 1 + e ^{u}} d u

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

Apply u-substitution

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

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

Factor

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

- \frac{1}{2} \cdot 2 (- (\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

(a^{2} - x^{2})^{\frac{1}{2}}

x^{2} + sin (x)

x^{3} + x + 1

\frac{x ^{3}}{( x - 1 )}

\frac{1}{( a - x ) ^{3}}

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