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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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 = 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

y^{^{'}} = a y + b e^{- b}

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

y = \frac{4}{( 7 x ) ^{7}}

\int \frac{1}{x ^{2} 64 x ^{2} + 25} d x

x y^{^{'}} + y - 3 = 0

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