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

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

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integral of e^{1/2 x^2}*x^3

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

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

Solution steps

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

Simplify

\frac{1}{2} x^{2} : \frac{x ^{2}}{2}

= \int e^{\frac{x ^{2}}{2}} x^{3} d x

Apply u-substitution

: \int \frac{e ^{\frac{u}{2}} u}{2} d u

= \int \frac{e ^{\frac{u}{2}} u}{2} d u

Take the constant out:

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

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

Apply u-substitution

: \int 4 e^{v} v d v

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

Take the constant out:

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

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

Apply Integration By Parts

: e^{v} v - \int e^{v} d v

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

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} ((4 + x^{2})^{- 1})

\int (x^{8} - \frac{2 x ^{5}}{7} - 5) d x

y^{^{'}} = \frac{x ^{2} + x y + y ^{2}}{x ^{2}}

d y = (x - 2 x y) d x

x^{2} + 4 x - 8

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