What is the integral of (7x)/(e^{x^2)} ?

The integral of (7x)/(e^{x^2)} is -7/2 e^{-x^2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (7x)/(e^{x^2)}

\int \frac{7 x}{e ^{x^{2}}} d x

- \frac{7}{2} e^{- x^{2}} + C

Solution steps

\int \frac{7 x}{e ^{x^{2}}} d x

Take the constant out:

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

= 7 \cdot \int \frac{x}{e ^{x^{2}}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{x}{e ^{x^{2}}} = (x) e^{- x^{2}}

= 7 \cdot \int x e^{- x^{2}} d x

Apply u-substitution

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

= 7 \cdot \int - \frac{e ^{u}}{2} d u

Take the constant out:

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

= 7 (- \frac{1}{2} \cdot \int e^{u} d u)

Use the common integral:

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

= 7 (- \frac{1}{2} e^{u})

Substitute back

u = - x^{2}

= 7 (- \frac{1}{2} e^{- x^{2}})

Simplify

7 (- \frac{1}{2} e^{- x^{2}}) : - \frac{7}{2} e^{- x^{2}}

= - \frac{7}{2} e^{- x^{2}}

Add a constant to the solution

= - \frac{7}{2} e^{- x^{2}} + C

\int \frac{1}{x ^{3}} + \frac{6}{x ^{7}} d x

n = 0 \sum \infty \frac{1}{( 2 + 2 n ) ^{3}}

\int_{- 1}^{3} 2 x^{2} + 8 d x

\int 4 y^{5} d y

x (x - 8)

What is the integral of (7x)/(e^{x^2)} ?
The integral of (7x)/(e^{x^2)} is -7/2 e^{-x^2}+C