What is the integral of e^{3x+9} ?

The integral of e^{3x+9} is 1/3 e^{3x+9}+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+9}

\int e^{3 x + 9} d x

\frac{1}{3} e^{3 x + 9} + C

Solution steps

\int e^{3 x + 9} d x

Apply u-substitution

: \int e^{u} \frac{1}{3} d u

= \int e^{u} \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{3} e^{u}

Substitute back

u = 3 x + 9

= \frac{1}{3} e^{3 x + 9}

Add a constant to the solution

= \frac{1}{3} e^{3 x + 9} + C

7 sin (4 t + 8) + 3 cos (4 t - 2)

x \to 1 lim (1 - x - [x])

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

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

2 x sin (x^{2}) - cos (x) - e^{- x - 2}