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

The integral of e^{3+3y} is 1/3 e^{3+3y}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of e^{3+3y}

\int e^{3 + 3 y} d y

\frac{1}{3} e^{3 + 3 y} + C

Solution steps

\int e^{3 + 3 y} d y

Apply u-substitution

= \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 + 3 y

= \frac{1}{3} e^{3 + 3 y}

Add a constant to the solution

= \frac{1}{3} e^{3 + 3 y} + C