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

The integral of x^2e^{3x^3-1} is 1/9 e^{3x^3-1}+C

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

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

\frac{1}{9} e^{3 x^{3} - 1} + C

Solution steps

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

x^{2} e^{3 x^{3} - 1} = e^{- 1} x^{2} e^{3 x^{3}}

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

Apply exponent rule:

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

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

Apply u-substitution

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

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

Apply u-substitution

: \int e^{v} \frac{1}{3} d v

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

Use the common integral:

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

= \frac{1}{e} \cdot \frac{1}{3} \cdot \frac{1}{3} e^{v}

Substitute back

= \frac{1}{e} \cdot \frac{1}{3} \cdot \frac{1}{3} e^{3 x^{3}}

Simplify

\frac{1}{e} \cdot \frac{1}{3} \cdot \frac{1}{3} e^{3 x^{3}} : \frac{1}{9} e^{3 x^{3} - 1}

= \frac{1}{9} e^{3 x^{3} - 1}

Add a constant to the solution

= \frac{1}{9} e^{3 x^{3} - 1} + C

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What is the integral of x^2e^{3x^3-1} ?
The integral of x^2e^{3x^3-1} is 1/9 e^{3x^3-1}+C