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

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

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

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

- \frac{1}{6} (3 - x^{2})^{3} + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

= - \frac{1}{2} \cdot \frac{u ^{3}}{3}

Substitute back

u = 3 - x^{2}

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

Simplify

- \frac{1}{2} \cdot \frac{( 3 - x ^{2} ) ^{3}}{3} : - \frac{1}{6} (3 - x^{2})^{3}

= - \frac{1}{6} (3 - x^{2})^{3}

Add a constant to the solution

= - \frac{1}{6} (3 - x^{2})^{3} + C

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