What is the integral of (4x^2-4)(e^{x^3-3x}) ?

The integral of (4x^2-4)(e^{x^3-3x}) is 4/3 e^{x^3-3x}+C

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integral of (4x^2-4)(e^{x^3-3x})

\int (4 x^{2} - 4) (e^{x^{3} - 3 x}) d x

\frac{4}{3} e^{x^{3} - 3 x} + C

Solution steps

\int (4 x^{2} - 4) e^{x^{3} - 3 x} d x

(4 x^{2} - 4) e^{x^{3} - 3 x} = 4 (x^{2} - 1) e^{x^{3} - 3 x}

= \int 4 (x^{2} - 1) e^{x^{3} - 3 x} d x

Take the constant out:

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

= 4 \cdot \int (x^{2} - 1) e^{x^{3} - 3 x} d x

Apply u-substitution

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

= 4 \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

= 4 \cdot \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= 4 \cdot \frac{1}{3} e^{u}

Substitute back

u = x^{3} - 3 x

= 4 \cdot \frac{1}{3} e^{x^{3} - 3 x}

Simplify

4 \cdot \frac{1}{3} e^{x^{3} - 3 x} : \frac{4}{3} e^{x^{3} - 3 x}

= \frac{4}{3} e^{x^{3} - 3 x}

Add a constant to the solution

= \frac{4}{3} e^{x^{3} - 3 x} + C

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What is the integral of (4x^2-4)(e^{x^3-3x}) ?
The integral of (4x^2-4)(e^{x^3-3x}) is 4/3 e^{x^3-3x}+C