What is the integral of e^{9x+4} ?

The integral of e^{9x+4} is 1/9 e^{9x+4}+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+4}

\int e^{9 x + 4} d x

\frac{1}{9} e^{9 x + 4} + C

Solution steps

\int e^{9 x + 4} 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 + 4

= \frac{1}{9} e^{9 x + 4}

Add a constant to the solution

= \frac{1}{9} e^{9 x + 4} + C

\int 1 + 3 x^{3} (9 x^{2}) d x

\int (4 + e^{x})^{2} e^{x} d x

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

\int - sin (x) + e^{x} + 3 x^{2} d x

\int x y^{2} z^{2} d x