What is the integral of (28e^{4x})/(sqrt(6+e^{4x))} ?

The integral of (28e^{4x})/(sqrt(6+e^{4x))} is 14sqrt(6+e^{4x)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (28e^{4x})/(sqrt(6+e^{4x))}

\int \frac{28 e ^{4 x}}{6 + e ^{4 x}} d x

14 6 + e^{4 x} + C

Solution steps

\int \frac{28 e ^{4 x}}{6 + e ^{4 x}} d x

Take the constant out:

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

= 28 \cdot \int \frac{e ^{4 x}}{6 + e ^{4 x}} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

= 28 \cdot \int \frac{1}{4 u} d u

Take the constant out:

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

= 28 \cdot \frac{1}{4} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= 28 \cdot \frac{1}{4} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 6 + e^{4 x}

= 28 \cdot \frac{1}{4} \cdot 2 (6 + e^{4 x})^{\frac{1}{2}}

Simplify

28 \cdot \frac{1}{4} \cdot 2 (6 + e^{4 x})^{\frac{1}{2}} : 14 6 + e^{4 x}

= 14 6 + e^{4 x}

Add a constant to the solution

= 14 6 + e^{4 x} + C

t e^{10 t}

x \to 0 lim (\frac{( arctan ( 4 x ) )}{ln ( x )})

\frac{\partial}{\partial x} (ln (4 x + 5 y))

x \to 0 + lim (e^{- 2 x} ln (x))

\frac{1}{x}, 3

What is the integral of (28e^{4x})/(sqrt(6+e^{4x))} ?
The integral of (28e^{4x})/(sqrt(6+e^{4x))} is 14sqrt(6+e^{4x)}+C