What is the integral of 1/(2+e^{4x)} ?

The integral of 1/(2+e^{4x)} is -1/8 (ln|e^{4x}+2|-4x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(2+e^{4x)}

\int \frac{1}{2 + e ^{4 x}} d x

- \frac{1}{8} (ln e^{4 x} + 2 - 4 x) + C

Solution steps

\int \frac{1}{2 + e ^{4 x}} d x

Apply u-substitution

: \int \frac{1}{4 u ( u - 2 )} d u

= \int \frac{1}{4 u ( u - 2 )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u ( u - 2 )} d u

Complete the square

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

= \frac{1}{4} \cdot \int \frac{1}{( u - 1 ) ^{2} - 1} d u

Apply u-substitution

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

= \frac{1}{4} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

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

= \frac{1}{4} \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}{4} (- \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}{4} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= \frac{1}{4} (- (\frac{ln 2 + e ^{4 x} - 1 + 1}{2} - \frac{ln 2 + e ^{4 x} - 1 - 1}{2}))

Simplify

\frac{1}{4} (- (\frac{ln 2 + e ^{4 x} - 1 + 1}{2} - \frac{ln 2 + e ^{4 x} - 1 - 1}{2})) : - \frac{1}{8} (ln e^{4 x} + 2 - 4 x)

= - \frac{1}{8} (ln e^{4 x} + 2 - 4 x)

Add a constant to the solution

= - \frac{1}{8} (ln e^{4 x} + 2 - 4 x) + C

\int \frac{sin ( x )}{6 + cos ( x )} d x

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

\int \frac{60}{x ^{2}} d x

\int cos (\frac{2 π x}{a}) d x

\int \frac{- 5 ln ( x )}{x} d x

What is the integral of 1/(2+e^{4x)} ?
The integral of 1/(2+e^{4x)} is -1/8 (ln|e^{4x}+2|-4x)+C