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

The integral of (x^3e^{-x^2}) is 1/2 (-e^{-x^2}x^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 (x^3e^{-x^2})

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

\frac{1}{2} (- e^{- x^{2}} x^{2} - e^{- x^{2}}) + C

Solution steps

\int x^{3} e^{- x^{2}} d x

Apply u-substitution

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

= \int \frac{e ^{u} u}{2} d u

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = - x^{2}

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

Simplify

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

Add a constant to the solution

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

\int sin (ln (3 x)) d x

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

\int_{0}^{l n (2)} e^{2 x} d x

\int \frac{9}{2} (t + 1)^{\frac{1}{2}} d t

y^{^{'}} = x^{3} (1 - y), y (0) = 3

What is the integral of (x^3e^{-x^2}) ?
The integral of (x^3e^{-x^2}) is 1/2 (-e^{-x^2}x^2-e^{-x^2})+C