What is the integral of p^3e^{p^2} ?

The integral of p^3e^{p^2} is 1/2 (e^{p^2}p^2-e^{p^2})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of p^3e^{p^2}

\int p^{3} e^{p^{2}} d p

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

Solution steps

\int p^{3} e^{p^{2}} d p

Apply u-substitution

: \int \frac{e ^{u^{\frac{2}{3}}} u ^{\frac{1}{3}}}{3} d u

= \int \frac{e ^{u^{\frac{2}{3}}} u ^{\frac{1}{3}}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int e^{u^{\frac{2}{3}}} u^{\frac{1}{3}} d u

Apply u-substitution

: \int \frac{3 e ^{v} v}{2} d v

= \frac{1}{3} \cdot \int \frac{3 e ^{v} v}{2} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{3}{2} \cdot \int e^{v} v d v

Apply Integration By Parts

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

= \frac{1}{3} \cdot \frac{3}{2} (e^{v} v - \int e^{v} d v)

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

= \frac{1}{3} \cdot \frac{3}{2} (e^{v} v - e^{v})

Substitute back

= \frac{1}{3} \cdot \frac{3}{2} (e^{(p^{3})^{\frac{2}{3}}} (p^{3})^{\frac{2}{3}} - e^{(p^{3})^{\frac{2}{3}}})

Simplify

\frac{1}{3} \cdot \frac{3}{2} (e^{(p^{3})^{\frac{2}{3}}} (p^{3})^{\frac{2}{3}} - e^{(p^{3})^{\frac{2}{3}}}) : \frac{1}{2} (e^{p^{2}} p^{2} - e^{p^{2}})

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

Add a constant to the solution

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

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

\int 28 x^{- \frac{1}{5}} tan (x^{\frac{4}{5}}) d x

\int x (1 + x) d x

\int 11 t d t

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