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

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

\int e^{2 x} (e^{2 x} - e^{- 2 x}) d x

\frac{1}{4} e^{4 x} - x + C

Solution steps

\int e^{2 x} (e^{2 x} - e^{- 2 x}) d x

Expand

e^{2 x} (e^{2 x} - e^{- 2 x}) : e^{4 x} - 1

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

Apply the Sum Rule:

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

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

\int e^{4 x} d x = \frac{1}{4} e^{4 x}

\int 1 d x = x

= \frac{1}{4} e^{4 x} - x

Add a constant to the solution

= \frac{1}{4} e^{4 x} - x + C

\frac{1}{1 - t}

\frac{d}{d x} ((4 + 2 x^{2})^{7})

\int (4 sin (2 x) - 3 cos (3 x)) d x

(2 + x)^{- 2}

x \to \infty lim (\frac{2 ( x - 8 ) ^{35}}{( x - 1 ) ^{35}})

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