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

The integral of (ar)/(sqrt(a^2-r^2)) is -asqrt(a^2-r^2)+C

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integral of (ar)/(sqrt(a^2-r^2))

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

- a a^{2} - r^{2} + C

Solution steps

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

Take the constant out:

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

= a \cdot \int \frac{r}{a ^{2} - r ^{2}} d r

Apply u-substitution

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

= a \cdot \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

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

Apply the Power Rule

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

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

Substitute back

u = a^{2} - r^{2}

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

Simplify

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

= - a a^{2} - r^{2}

Add a constant to the solution

= - a a^{2} - r^{2} + C

\frac{d}{d x} (\frac{2}{2})

\int \frac{e ^{x}}{e ^{2 x} + 3 e ^{x} + 2} d x

\int 5 e^{5 x} d x

f (x) = - 6 x^{2} - 9, at x = 0

x \to - 4 - lim (f (x))

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