What is the integral of r/(sqrt(4-r^2)) ?

The integral of r/(sqrt(4-r^2)) is -sqrt(4-r^2)+C

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integral of r/(sqrt(4-r^2))

\int \frac{r}{4 - r ^{2}} d r

- 4 - r^{2} + C

Solution steps

\int \frac{r}{4 - r ^{2}} d r

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 = 4 - r^{2}

= - \frac{1}{2} \cdot 2 (4 - r^{2})^{\frac{1}{2}}

Simplify

- \frac{1}{2} \cdot 2 (4 - r^{2})^{\frac{1}{2}} : - 4 - r^{2}

= - 4 - r^{2}

Add a constant to the solution

= - 4 - r^{2} + C

\int (5 x - 1)^{3} d x

\frac{3 x ^{2}}{( 3 - 2 x ) ^{4}}

y^{^{''}} + 16 y = 2 sin (2 t), y (0) = 1, y^{^{'}} (0) = 0

\int (2 t e^{t}) d t

y^{4} - 4 y^{2} = x^{4} - 9 x^{2}, (- 3, 2)

What is the integral of r/(sqrt(4-r^2)) ?
The integral of r/(sqrt(4-r^2)) is -sqrt(4-r^2)+C