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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

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^{(t^{3})^{\frac{2}{3}}} (t^{3})^{\frac{2}{3}} - e^{(t^{3})^{\frac{2}{3}}})

Simplify

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

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

Add a constant to the solution

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

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

\frac{d y}{d x}, 15 x = 15 y + 5 y^{3} + 3 y^{5}

y^{^{'}} + \frac{y}{x} = 5 \cdot x^{3}

\frac{d}{d x} (e^{- 2 x} \cdot sin (x))

\int 1 + e^{2 x} d x

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