What is the integral of e^{3x}3^x ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{3x}3^x

\int e^{3 x} 3^{x} d x

\frac{1}{3 + ln ( 3 )} \cdot 3^{\frac{( 3 + l n ( 3 ) ) x}{l n ( 3 )}} + C

Solution steps

\int e^{3 x} \cdot 3^{x} d x

Apply u-substitution

: \int \frac{u ^{\frac{3}{l n ( 3 )}}}{ln ( 3 )} d u

= \int \frac{u ^{\frac{3}{l n ( 3 )}}}{ln ( 3 )} d u

Take the constant out:

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

= \frac{1}{ln ( 3 )} \cdot \int u^{\frac{3}{l n ( 3 )}} d u

Apply the Power Rule

: \frac{1}{3 + ln ( 3 )} ln (3) u^{\frac{3 + l n ( 3 )}{l n ( 3 )}}

= \frac{1}{ln ( 3 )} \cdot \frac{1}{3 + ln ( 3 )} ln (3) u^{\frac{3 + l n ( 3 )}{l n ( 3 )}}

Substitute back

u = 3^{x}

= \frac{1}{ln ( 3 )} \cdot \frac{1}{3 + ln ( 3 )} ln (3) (3^{x})^{\frac{3 + l n ( 3 )}{l n ( 3 )}}

Simplify

\frac{1}{ln ( 3 )} \cdot \frac{1}{3 + ln ( 3 )} ln (3) (3^{x})^{\frac{3 + l n ( 3 )}{l n ( 3 )}} : \frac{1}{3 + ln ( 3 )} \cdot 3^{\frac{( 3 + l n ( 3 ) ) x}{l n ( 3 )}}

= \frac{1}{3 + ln ( 3 )} \cdot 3^{\frac{( 3 + l n ( 3 ) ) x}{l n ( 3 )}}

Add a constant to the solution

= \frac{1}{3 + ln ( 3 )} \cdot 3^{\frac{( 3 + l n ( 3 ) ) x}{l n ( 3 )}} + C

\int \frac{1}{2 + x ^{3}} d x

\int (1 + tan^{2} (β)) d β

\int x (x^{2} + 2019)^{2020} d x

\int \frac{sin ( x )}{cos ^{2} ( x ) + 4} d x

\int x^{- 1} (x^{- 2} + x^{- 1} + 2 + x^{2}) d x

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