What is the integral of sqrt(-3+r^2)r ?

The integral of sqrt(-3+r^2)r is 1/3 (-3+r^2)^{3/2}+C

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integral of sqrt(-3+r^2)r

\int - 3 + r^{2} r d r

\frac{1}{3} (- 3 + r^{2})^{\frac{3}{2}} + C

Solution steps

\int - 3 + r^{2} r d r

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{u ^{3}}{3}

Substitute back

u = - 3 + r^{2}

= \frac{( - 3 + r ^{2} ) ^{3}}{3}

Simplify

\frac{( - 3 + r ^{2} ) ^{3}}{3} : \frac{1}{3} (- 3 + r^{2})^{\frac{3}{2}}

= \frac{1}{3} (- 3 + r^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} (- 3 + r^{2})^{\frac{3}{2}} + C

T + 2

y^{^{'}} = - y + 5

x \to 1 lim (\frac{x - 3}{2})

\frac{d}{d x} (arctan (5 cos (x)))

k = 1 \sum \infty \frac{1}{4 ^{k}}

What is the integral of sqrt(-3+r^2)r ?
The integral of sqrt(-3+r^2)r is 1/3 (-3+r^2)^{3/2}+C