What is the integral of 6e^{2x+1} ?

The integral of 6e^{2x+1} is 3e^{2x+1}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6e^{2x+1}

\int 6 e^{2 x + 1} d x

3 e^{2 x + 1} + C

Solution steps

\int 6 e^{2 x + 1} d x

Take the constant out:

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

= 6 \cdot \int e^{2 x + 1} d x

Apply u-substitution

: \int e^{u} \frac{1}{2} d u

= 6 \cdot \int e^{u} \frac{1}{2} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= 6 \cdot \frac{1}{2} e^{u}

Substitute back

u = 2 x + 1

= 6 \cdot \frac{1}{2} e^{2 x + 1}

Simplify

6 \cdot \frac{1}{2} e^{2 x + 1} : 3 e^{2 x + 1}

= 3 e^{2 x + 1}

Add a constant to the solution

= 3 e^{2 x + 1} + C

\int 9 cos^{2} (t) d t

\int \frac{x + 4}{x ^{2} + 3 x + 2} d x

\int 3 t - 8 d t

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

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