What is the integral of 3^{2x-1} ?

The integral of 3^{2x-1} is (3^{2x-1})/(2ln(3))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 3^{2x-1}

\int 3^{2 x - 1} d x

\frac{3 ^{2 x - 1}}{2 ln ( 3 )} + C

Solution steps

\int 3^{2 x - 1} d x

Apply u-substitution

= \int 3^{u} \cdot \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 3^{u} d u

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

= \frac{1}{2} \cdot \frac{3 ^{u}}{ln ( 3 )}

Substitute back

u = 2 x - 1

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

Simplify

\frac{1}{2} \cdot \frac{3 ^{2 x - 1}}{ln ( 3 )} : \frac{3 ^{2 x - 1}}{2 ln ( 3 )}

= \frac{3 ^{2 x - 1}}{2 ln ( 3 )}

Add a constant to the solution

= \frac{3 ^{2 x - 1}}{2 ln ( 3 )} + C

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