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

The integral of 2y^3e^{y^2} is e^{y^2}y^2-e^{y^2}+C

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

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

e^{y^{2}} y^{2} - e^{y^{2}} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 2 \cdot \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

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

Apply u-substitution

= 2 \cdot \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

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

Apply Integration By Parts

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

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

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

Substitute back

= 2 \cdot \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

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

= e^{y^{2}} y^{2} - e^{y^{2}}

Add a constant to the solution

= e^{y^{2}} y^{2} - e^{y^{2}} + C

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