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

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

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integral of w/(sqrt(1-w^2))

\int \frac{w}{1 - w ^{2}} d w

- 1 - w^{2} + C

Solution steps

\int \frac{w}{1 - w ^{2}} d w

Apply u-substitution

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

= \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

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

Apply the Power Rule

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

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

Substitute back

u = 1 - w^{2}

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

Simplify

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

= - 1 - w^{2}

Add a constant to the solution

= - 1 - w^{2} + C

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