What is the integral of x^2e^{x^3-5} ?

The integral of x^2e^{x^3-5} is 1/3 e^{x^3-5}+C

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integral of x^2e^{x^3-5}

\int x^{2} e^{x^{3} - 5} d x

\frac{1}{3} e^{x^{3} - 5} + C

Solution steps

\int x^{2} e^{x^{3} - 5} d x

x^{2} e^{x^{3} - 5} = e^{- 5} x^{2} e^{x^{3}}

= \int e^{- 5} x^{2} e^{x^{3}} d x

Take the constant out:

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

= e^{- 5} \cdot \int x^{2} e^{x^{3}} d x

Apply exponent rule:

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

= \frac{1}{e ^{5}} \cdot \int x^{2} e^{x^{3}} d x

Apply u-substitution

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

= \frac{1}{e ^{5}} \cdot \int \frac{e ^{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 ^{5}} \cdot \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{e ^{5}} \cdot \frac{1}{3} e^{u}

Substitute back

u = x^{3}

= \frac{1}{e ^{5}} \cdot \frac{1}{3} e^{x^{3}}

Simplify

\frac{1}{e ^{5}} \cdot \frac{1}{3} e^{x^{3}} : \frac{1}{3} e^{x^{3} - 5}

= \frac{1}{3} e^{x^{3} - 5}

Add a constant to the solution

= \frac{1}{3} e^{x^{3} - 5} + C

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What is the integral of x^2e^{x^3-5} ?
The integral of x^2e^{x^3-5} is 1/3 e^{x^3-5}+C