What is the integral of (6x)/(sqrt(1-x^2)) ?

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

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integral of (6x)/(sqrt(1-x^2))

\int \frac{6 x}{1 - x ^{2}} d x

- 6 1 - x^{2} + C

Solution steps

\int \frac{6 x}{1 - x ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{x}{1 - x ^{2}} d x

Apply u-substitution

: \int - \frac{1}{2 u} d u

= 6 \cdot \int - \frac{1}{2 u} d u

Take the constant out:

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

= 6 (- \frac{1}{2} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= 6 (- \frac{1}{2} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 1 - x^{2}

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

Simplify

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

= - 6 1 - x^{2}

Add a constant to the solution

= - 6 1 - x^{2} + C

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