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

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

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integral of xf(1-x^2)

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

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

Solution steps

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

Take the constant out:

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

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

Expand

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

= f \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

= f (\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}

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

Add a constant to the solution

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

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