What is the integral of e^{(x^3-6x)}(x^2-2) ?

The integral of e^{(x^3-6x)}(x^2-2) is 1/3 e^{x^3-6x}+C

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integral of e^{(x^3-6x)}(x^2-2)

\int e^{(x^{3} - 6 x)} (x^{2} - 2) d x

\frac{1}{3} e^{x^{3} - 6 x} + C

Solution steps

\int e^{x^{3} - 6 x} (x^{2} - 2) d x

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = x^{3} - 6 x

= \frac{1}{3} e^{x^{3} - 6 x}

Add a constant to the solution

= \frac{1}{3} e^{x^{3} - 6 x} + C

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What is the integral of e^{(x^3-6x)}(x^2-2) ?
The integral of e^{(x^3-6x)}(x^2-2) is 1/3 e^{x^3-6x}+C