What is the integral of 2βx(1-x^2)^{β-1} ?

The integral of 2βx(1-x^2)^{β-1} is -(1-x^2)^β+C

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integral of 2βx(1-x^2)^{β-1}

\int 2 β x (1 - x^{2})^{β - 1} d x

- (1 - x^{2})^{β} + C

Solution steps

\int 2 β x (1 - x^{2})^{β - 1} d x

Take the constant out:

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

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

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

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

Substitute back

u = 1 - x^{2}

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

Simplify

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

= - (1 - x^{2})^{β}

Add a constant to the solution

= - (1 - x^{2})^{β} + C

What is the integral of 2βx(1-x^2)^{β-1} ?
The integral of 2βx(1-x^2)^{β-1} is -(1-x^2)^β+C