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

The integral of e^{2x}sqrt(1-4e^{2x)} is -1/12 (1-4e^{2x})^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{2x}sqrt(1-4e^{2x)}

\int e^{2 x} 1 - 4 e^{2 x} d x

- \frac{1}{12} (1 - 4 e^{2 x})^{\frac{3}{2}} + C

Solution steps

\int e^{2 x} 1 - 4 e^{2 x} d x

Apply u-substitution

: \int e^{u} 1 - 4 e^{u} \frac{1}{2} d u

= \int e^{u} 1 - 4 e^{u} \frac{1}{2} 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 e^{u} 1 - 4 e^{u} d u

Apply u-substitution

: \int 1 - 4 v d v

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

Apply u-substitution

: \int - \frac{w}{4} d w

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{2}{3} w^{\frac{3}{2}}

= \frac{1}{2} (- \frac{1}{4} \cdot \frac{2}{3} w^{\frac{3}{2}})

Substitute back

= \frac{1}{2} (- \frac{1}{4} \cdot \frac{2}{3} (1 - 4 e^{2 x})^{\frac{3}{2}})

Simplify

\frac{1}{2} (- \frac{1}{4} \cdot \frac{2}{3} (1 - 4 e^{2 x})^{\frac{3}{2}}) : - \frac{1}{12} (1 - 4 e^{2 x})^{\frac{3}{2}}

= - \frac{1}{12} (1 - 4 e^{2 x})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{12} (1 - 4 e^{2 x})^{\frac{3}{2}} + C

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

x \to \infty lim (\frac{4 x - 5}{2 x + 3})

n = 1 \sum \infty \frac{4 n}{4 n + 7}

(2 x + 5) (4 x - 5)^{2}

\frac{d}{d x} (ln (x^{s i n (x)}))

What is the integral of e^{2x}sqrt(1-4e^{2x)} ?
The integral of e^{2x}sqrt(1-4e^{2x)} is -1/12 (1-4e^{2x})^{3/2}+C