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

The integral of sqrt(4+2r^2)r is 1/6 (4+2r^2)^{3/2}+C

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integral of sqrt(4+2r^2)r

\int 4 + 2 r^{2} r d r

\frac{1}{6} (4 + 2 r^{2})^{\frac{3}{2}} + C

Solution steps

\int 4 + 2 r^{2} r d r

Apply u-substitution

: \int \frac{u}{4} d u

= \int \frac{u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 + 2 r^{2}

= \frac{1}{4} \cdot \frac{2}{3} (4 + 2 r^{2})^{\frac{3}{2}}

Simplify

\frac{1}{4} \cdot \frac{2}{3} (4 + 2 r^{2})^{\frac{3}{2}} : \frac{1}{6} (4 + 2 r^{2})^{\frac{3}{2}}

= \frac{1}{6} (4 + 2 r^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{6} (4 + 2 r^{2})^{\frac{3}{2}} + C

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What is the integral of sqrt(4+2r^2)r ?
The integral of sqrt(4+2r^2)r is 1/6 (4+2r^2)^{3/2}+C