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

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

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

- ln e^{x} + 1 + 1 + ln e^{x} + 1 - 1 + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Factor

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

= 2 \cdot \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

= 2 (- \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}

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

Substitute back

u = e^{x} + 1

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

Simplify

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

= - ln e^{x} + 1 + 1 + ln e^{x} + 1 - 1

Add a constant to the solution

= - ln e^{x} + 1 + 1 + ln e^{x} + 1 - 1 + C

n = 0 \sum \infty \frac{1}{( 2 n + 1 ) ^{3}}

n \to \infty lim (- 7^{n})

\frac{1}{s ( s ^{2} + 100 )}

\frac{\partial}{\partial x} (x y e^{7 y})

y = ln (x^{3} + 1)

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