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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

n = 5 \sum \infty \frac{9 ^{n}}{1 0 ^{n}}

\frac{d}{d x} (sin (x) - \frac{1}{3} sin^{3} (x))

\int - 25 sin (5 t - \frac{π}{2}) d t

y^{^{'}} + x e^{y} = 0

\int e^{x} 5 + e^{x} d x

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