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

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

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

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

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

Solution steps

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

Simplify

e^{x^{2} - 1} : e^{x^{2}} e^{- 1}

= \int x^{3} e^{x^{2}} e^{- 1} d x

Take the constant out:

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

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

Apply u-substitution

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

= \frac{1}{e} \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

= \frac{1}{e} \cdot \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}{e} \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

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

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

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

Substitute back

= \frac{1}{e} \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

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

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

Add a constant to the solution

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

x \to \infty lim (\frac{4}{3})

y = x, (16, 4)

y^{^{'}} + \frac{y}{x} = \frac{7}{x ^{2}} + 3

\int (e x^{2} + π) d x

x \to 0 lim (\frac{( 2 ∣ x ∣ )}{3 x})

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