What is the integral of 2x(3-x^2)^3 ?

The integral of 2x(3-x^2)^3 is -1/4 (3-x^2)^4+C

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integral of 2x(3-x^2)^3

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

- \frac{1}{4} (3 - x^{2})^{4} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int - \frac{u ^{3}}{2} d u

= 2 \cdot \int - \frac{u ^{3}}{2} d u

Take the constant out:

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

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

Apply the Power Rule

: \frac{u ^{4}}{4}

= 2 (- \frac{1}{2} \cdot \frac{u ^{4}}{4})

Substitute back

u = 3 - x^{2}

= 2 (- \frac{1}{2} \cdot \frac{( 3 - x ^{2} ) ^{4}}{4})

Simplify

2 (- \frac{1}{2} \cdot \frac{( 3 - x ^{2} ) ^{4}}{4}) : - \frac{1}{4} (3 - x^{2})^{4}

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

Add a constant to the solution

= - \frac{1}{4} (3 - x^{2})^{4} + C

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What is the integral of 2x(3-x^2)^3 ?
The integral of 2x(3-x^2)^3 is -1/4 (3-x^2)^4+C