What is the integral of e^{9x+1} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of e^{9x+1}

\int e^{9 x + 1} d x

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

Solution steps

\int e^{9 x + 1} d x

Apply u-substitution

= \int e^{u} \frac{1}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{9} e^{u}

Substitute back

u = 9 x + 1

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

Add a constant to the solution

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