What is the integral of r/(sqrt(16-9r^2)) ?

The integral of r/(sqrt(16-9r^2)) is -1/9 sqrt(16-9r^2)+C

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integral of r/(sqrt(16-9r^2))

\int \frac{r}{16 - 9 r ^{2}} d r

- \frac{1}{9} 16 - 9 r^{2} + C

Solution steps

\int \frac{r}{16 - 9 r ^{2}} d r

Apply u-substitution

: \int - \frac{1}{18 u} d u

= \int - \frac{1}{18 u} d u

Take the constant out:

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

= - \frac{1}{18} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= - \frac{1}{18} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 16 - 9 r^{2}

= - \frac{1}{18} \cdot 2 (16 - 9 r^{2})^{\frac{1}{2}}

Simplify

- \frac{1}{18} \cdot 2 (16 - 9 r^{2})^{\frac{1}{2}} : - \frac{1}{9} 16 - 9 r^{2}

= - \frac{1}{9} 16 - 9 r^{2}

Add a constant to the solution

= - \frac{1}{9} 16 - 9 r^{2} + C

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What is the integral of r/(sqrt(16-9r^2)) ?
The integral of r/(sqrt(16-9r^2)) is -1/9 sqrt(16-9r^2)+C