What is the integral of sqrt(4r^2+1)r ?

The integral of sqrt(4r^2+1)r is 1/12 (4r^2+1)^{3/2}+C

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integral of sqrt(4r^2+1)r

\int 4 r^{2} + 1 r d r

\frac{1}{12} (4 r^{2} + 1)^{\frac{3}{2}} + C

Solution steps

\int 4 r^{2} + 1 r d r

Apply u-substitution

: \int \frac{u}{8} d u

= \int \frac{u}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{8} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 r^{2} + 1

= \frac{1}{8} \cdot \frac{2}{3} (4 r^{2} + 1)^{\frac{3}{2}}

Simplify

\frac{1}{8} \cdot \frac{2}{3} (4 r^{2} + 1)^{\frac{3}{2}} : \frac{1}{12} (4 r^{2} + 1)^{\frac{3}{2}}

= \frac{1}{12} (4 r^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{12} (4 r^{2} + 1)^{\frac{3}{2}} + C

\frac{d}{d x} (arccos (\frac{x}{8}))

\frac{d}{d x} (π^{x})

\int \frac{1}{x 9 x ^{2} - 25} d x

\frac{d y}{d x} = \frac{4}{x ^{2} e ^{2 y}}

\int sin (x) d x

What is the integral of sqrt(4r^2+1)r ?
The integral of sqrt(4r^2+1)r is 1/12 (4r^2+1)^{3/2}+C