What is the integral of 4x(1-x^2) ?

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

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integral of 4x(1-x^2)

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

2 x^{2} - x^{4} + C

Solution steps

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

Take the constant out:

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

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

Expand

x (1 - x^{2}) : x - x^{3}

= 4 \cdot \int x - x^{3} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int x d x = \frac{x ^{2}}{2}

\int x^{3} d x = \frac{x ^{4}}{4}

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

Simplify

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

= 2 x^{2} - x^{4}

Add a constant to the solution

= 2 x^{2} - x^{4} + C

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