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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{2} (u - ln ∣ u ∣)

Substitute back

u = e^{2 x} + 1

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

Add a constant to the solution

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

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

\int sin^{3} (m) t cos^{2} (m) t d t

\int \frac{1}{( arctan ( \frac{y}{2} ) ) ( 4 + y ^{2} )} d y

\int 1 sin (x) d x

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

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