What is the integral of (e^{8x}+4e^{-x}) ?

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

\int (e^{8 x} + 4 e^{- x}) d x

\frac{1}{8 e ^{- 8 x}} - 4 e^{- x} + C

Solution steps

\int e^{8 x} + 4 e^{- x} d x

Apply u-substitution

: \int - \frac{1 + 4 u ^{9}}{u ^{9}} d u

= \int - \frac{1 + 4 u ^{9}}{u ^{9}} d u

Take the constant out:

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

= - \int \frac{1 + 4 u ^{9}}{u ^{9}} d u

Expand

\frac{1 + 4 u ^{9}}{u ^{9}} : \frac{1}{u ^{9}} + 4

= - \int \frac{1}{u ^{9}} + 4 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

= - (\int \frac{1}{u ^{9}} d u + \int 4 d u)

\int \frac{1}{u ^{9}} d u = - \frac{1}{8 u ^{8}}

\int 4 d u = 4 u

= - (- \frac{1}{8 u ^{8}} + 4 u)

Substitute back

u = e^{- x}

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

Simplify

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

= \frac{1}{8 e ^{- 8 x}} - 4 e^{- x}

Add a constant to the solution

= \frac{1}{8 e ^{- 8 x}} - 4 e^{- x} + C

\int 7 x - 2 d x

x \to - 1 - lim (\frac{x}{x ^{2} + 1})

\int_{0}^{6} (6 x - x^{2}) d x

\frac{d}{d x} (\frac{x + x}{x})

x^{2} y^{^{'}} + 2 x y = ln (x), y (1) = 5

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