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

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

Popular >

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

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

- 1 - r^{2} + C

Solution steps

\int \frac{r}{1 - 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 = 1 - r^{2}

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

Simplify

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

= - 1 - r^{2}

Add a constant to the solution

= - 1 - r^{2} + C

\int (3 x^{\frac{1}{2}} - x^{\frac{- 2}{3}} + 7) d x

\int 0.25 D 100 - D d D

\int (4 + x^{- 4} - e^{3 x + 1}) d x

\int arcsin (x - 1) d x

\int (3 csc (3 x) + 2 cot (3 x))^{2} d x

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