What is the integral of 9xe^{-5x^2} ?

The integral of 9xe^{-5x^2} is -9/10 e^{-5x^2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 9xe^{-5x^2}

\int 9 x e^{- 5 x^{2}} d x

- \frac{9}{10} e^{- 5 x^{2}} + C

Solution steps

\int 9 x e^{- 5 x^{2}} d x

Take the constant out:

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

= 9 \cdot \int x e^{- 5 x^{2}} d x

Apply u-substitution

: \int - \frac{e ^{u}}{10} d u

= 9 \cdot \int - \frac{e ^{u}}{10} d u

Take the constant out:

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

= 9 (- \frac{1}{10} \cdot \int e^{u} d u)

Use the common integral:

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

= 9 (- \frac{1}{10} e^{u})

Substitute back

u = - 5 x^{2}

= 9 (- \frac{1}{10} e^{- 5 x^{2}})

Simplify

9 (- \frac{1}{10} e^{- 5 x^{2}}) : - \frac{9}{10} e^{- 5 x^{2}}

= - \frac{9}{10} e^{- 5 x^{2}}

Add a constant to the solution

= - \frac{9}{10} e^{- 5 x^{2}} + C

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

\int \frac{( x + 2 )}{( x ^{2} + 4 x + 5 )} d x

\int \frac{1}{( x - 1 ) ( 3 - x )} d x

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

\int \frac{4 x ^{3}}{x ^{2} - 16} d x

What is the integral of 9xe^{-5x^2} ?
The integral of 9xe^{-5x^2} is -9/10 e^{-5x^2}+C