What is the integral of (8+e^{-5x})/(e^{5x)} ?

The integral of (8+e^{-5x})/(e^{5x)} is -1/10 (8+e^{-5x})^2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (8+e^{-5x})/(e^{5x)}

\int \frac{8 + e ^{- 5 x}}{e ^{5 x}} d x

- \frac{1}{10} (8 + e^{- 5 x})^{2} + C

Solution steps

\int \frac{8 + e ^{- 5 x}}{e ^{5 x}} d x

Apply u-substitution

: \int - \frac{u}{5} d u

= \int - \frac{u}{5} d u

Take the constant out:

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

= - \frac{1}{5} \cdot \int u d u

Apply the Power Rule

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

= - \frac{1}{5} \cdot \frac{u ^{2}}{2}

Substitute back

u = 8 + e^{- 5 x}

= - \frac{1}{5} \cdot \frac{( 8 + e ^{- 5 x} ) ^{2}}{2}

Simplify

- \frac{1}{5} \cdot \frac{( 8 + e ^{- 5 x} ) ^{2}}{2} : - \frac{1}{10} (8 + e^{- 5 x})^{2}

= - \frac{1}{10} (8 + e^{- 5 x})^{2}

Add a constant to the solution

= - \frac{1}{10} (8 + e^{- 5 x})^{2} + C

\int sec^{2} (θ) e^{6 t a n (θ)} d θ

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

\int \frac{e ^{x}}{3 - 5 e ^{x}} d x

\int \frac{1}{1 - ln ( x )} d x

\int (3 x + 4)^{5} d x

What is the integral of (8+e^{-5x})/(e^{5x)} ?
The integral of (8+e^{-5x})/(e^{5x)} is -1/10 (8+e^{-5x})^2+C