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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 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 = 1 + e^{2 x}

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{( 3 x ^{2} - 2 )}{x ( x + 1 ) ^{3}} d x

s im pl i f y lo g_{10} (\frac{( x + 1 ) ( x - 8 )}{( x - 1 ) ( x + 8 )})

\int 3^{x} + \frac{2}{x} d x

\frac{\partial}{\partial x} (e^{8 x y})

\frac{d}{d x} (x^{k})

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