What is the integral of 5^{-2x+1} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 5^{-2x+1}

\int 5^{- 2 x + 1} d x

- \frac{5 ^{- 2 x + 1}}{2 ln ( 5 )} + C

Solution steps

\int 5^{- 2 x + 1} d x

Apply u-substitution

: \int - \frac{1}{2} \cdot 5^{u} d u

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

\int a^{x} d x = \frac{a ^{x}}{ln a}

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

Substitute back

u = - 2 x + 1

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

Simplify

- \frac{1}{2} \cdot \frac{5 ^{- 2 x + 1}}{ln ( 5 )} : - \frac{5 ^{- 2 x + 1}}{2 ln ( 5 )}

= - \frac{5 ^{- 2 x + 1}}{2 ln ( 5 )}

Add a constant to the solution

= - \frac{5 ^{- 2 x + 1}}{2 ln ( 5 )} + C

\int y^{2} e^{- 3 y} d y

\int x^{8} + \frac{1}{x ^{8}} d x

\int \frac{30}{36 + 25 x ^{2}} d x

\int \frac{1}{( 1 + cos ( 2 x ) )} d x

\int y e^{y - x} d y

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