What is the integral of (e^{9x}) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (e^{9x})

\int (e^{9 x}) d x

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

Solution steps

\int e^{9 x} d x

Apply u-substitution

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

= \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

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

Add a constant to the solution

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

\int x^{\frac{3}{4}} - \frac{1}{x ^{\frac{3}{4}}} d x

\int (x^{3} + 2 x + 1) d x

\int 7 x^{\frac{1}{2}} d x

\int (\frac{1}{7 x - 2} - \frac{1}{7 x + 2}) d x

\int sin (x) cos^{9} (x) d x