What is the integral of x^3e^{x^2+1} ?

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

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integral of x^3e^{x^2+1}

\int x^{3} e^{x^{2} + 1} d x

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

Solution steps

\int x^{3} e^{x^{2} + 1} d x

Apply u-substitution

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

Simplify

e^{u^{\frac{2}{3}} + 1} : e^{u^{\frac{2}{3}}} e

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

Take the constant out:

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

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

Apply u-substitution

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

Apply Integration By Parts

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

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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