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

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

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

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

- 1 - u^{2} + C

Solution steps

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

Apply u-substitution

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

= \int - \frac{1}{2 v} d v

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}{v} d v

Apply the Power Rule

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

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

Substitute back

v = 1 - u^{2}

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

Simplify

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

= - 1 - u^{2}

Add a constant to the solution

= - 1 - u^{2} + C

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\int \frac{x ^{2}}{25 + x ^{6}} d x

\frac{\partial}{\partial x} (e^{- 4 t} cos (π x))

y = \frac{x}{x + 4}, (1, 0.2)

\frac{d}{d y} (- 2 y)

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