What is the integral of 4x^3e^{x^2} ?

The integral of 4x^3e^{x^2} is 2(e^{x^2}x^2-e^{x^2})+C

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integral of 4x^3e^{x^2}

\int 4 x^{3} e^{x^{2}} d x

2 (e^{x^{2}} x^{2} - e^{x^{2}}) + C

Solution steps

\int 4 x^{3} e^{x^{2}} d x

Take the constant out:

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

= 4 \cdot \int x^{3} e^{x^{2}} d x

Apply u-substitution

= 4 \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

= 4 \cdot \frac{1}{3} \cdot \int e^{u^{\frac{2}{3}}} u^{\frac{1}{3}} d u

Apply u-substitution

= 4 \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

= 4 \cdot \frac{1}{3} \cdot \frac{3}{2} \cdot \int e^{v} v d v

Apply Integration By Parts

= 4 \cdot \frac{1}{3} \cdot \frac{3}{2} (e^{v} v - \int e^{v} d v)

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

= 4 \cdot \frac{1}{3} \cdot \frac{3}{2} (e^{v} v - e^{v})

Substitute back

= 4 \cdot \frac{1}{3} \cdot \frac{3}{2} (e^{(x^{3})^{\frac{2}{3}}} (x^{3})^{\frac{2}{3}} - e^{(x^{3})^{\frac{2}{3}}})

Simplify

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

= 2 (e^{x^{2}} x^{2} - e^{x^{2}})

Add a constant to the solution

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

What is the integral of 4x^3e^{x^2} ?
The integral of 4x^3e^{x^2} is 2(e^{x^2}x^2-e^{x^2})+C