What is the integral of 6(e^{4-3x})^2 ?

The integral of 6(e^{4-3x})^2 is -e^{8-6x}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6(e^{4-3x})^2

\int 6 (e^{4 - 3 x})^{2} d x

- e^{8 - 6 x} + C

Solution steps

\int 6 (e^{4 - 3 x})^{2} d x

Take the constant out:

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

= 6 \cdot \int (e^{4 - 3 x})^{2} d x

(e^{4 - 3 x})^{2} = (e^{4})^{2} (e^{- 3 x})^{2}

= 6 \cdot \int (e^{4})^{2} (e^{- 3 x})^{2} d x

Take the constant out:

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

= 6 (e^{4})^{2} \cdot \int (e^{- 3 x})^{2} d x

(e^{4})^{2} = e^{8}

= 6 e^{8} \cdot \int (e^{- 3 x})^{2} d x

Simplify

(e^{- 3 x})^{2} : e^{- 6 x}

= 6 e^{8} \cdot \int e^{- 6 x} d x

Apply u-substitution

= 6 e^{8} \cdot \int - \frac{1}{6} e^{u} d u

Take the constant out:

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

= 6 e^{8} (- \frac{1}{6} \cdot \int e^{u} d u)

Use the common integral:

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

= 6 e^{8} (- \frac{1}{6} e^{u})

Substitute back

u = - 6 x

= 6 e^{8} (- \frac{1}{6} e^{- 6 x})

Simplify

6 e^{8} (- \frac{1}{6} e^{- 6 x}) : - e^{8 - 6 x}

= - e^{8 - 6 x}

Add a constant to the solution

= - e^{8 - 6 x} + C

\int \frac{3 x + 1}{- 5 - 6 x - x ^{2}} d x

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

\int \frac{y}{( y + 5 ) ^{2} ( y - 5 )} d y

\int \frac{e ^{x}}{e ^{2 x} + 5 e ^{x} + 6} d x

\int x \cdot (1 - x) d x

What is the integral of 6(e^{4-3x})^2 ?
The integral of 6(e^{4-3x})^2 is -e^{8-6x}+C