What is the integral of r^3sqrt(1+4r^2) ?

The integral of r^3sqrt(1+4r^2) is 1/120 (6r^2-1)(4r^2+1)^{3/2}+C

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integral of r^3sqrt(1+4r^2)

\int r^{3} 1 + 4 r^{2} d r

\frac{1}{120} (6 r^{2} - 1) (4 r^{2} + 1)^{\frac{3}{2}} + C

Solution steps

\int r^{3} 1 + 4 r^{2} d r

Apply u-substitution

: \int \frac{u ( u - 1 )}{32} d u

= \int \frac{u ( u - 1 )}{32} d u

Take the constant out:

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

= \frac{1}{32} \cdot \int u (u - 1) d u

Expand

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

= \frac{1}{32} \cdot \int u^{\frac{3}{2}} - u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{32} (\int u^{\frac{3}{2}} d u - \int u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{32} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 + 4 r^{2}

= \frac{1}{32} (\frac{2}{5} (1 + 4 r^{2})^{\frac{5}{2}} - \frac{2}{3} (1 + 4 r^{2})^{\frac{3}{2}})

Simplify

\frac{1}{32} (\frac{2}{5} (1 + 4 r^{2})^{\frac{5}{2}} - \frac{2}{3} (1 + 4 r^{2})^{\frac{3}{2}}) : \frac{1}{120} (6 r^{2} - 1) (4 r^{2} + 1)^{\frac{3}{2}}

= \frac{1}{120} (6 r^{2} - 1) (4 r^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{120} (6 r^{2} - 1) (4 r^{2} + 1)^{\frac{3}{2}} + C

\frac{1}{3 ( 1 + e ^{- t} ) ^{3}}

x d y + 3 y d x = x y d y

x \to + 0 lim (\frac{- 6}{x ^{7}})

\int_{0}^{3} ∣ x - 2 ∣ d x

What is the integral of r^3sqrt(1+4r^2) ?
The integral of r^3sqrt(1+4r^2) is 1/120 (6r^2-1)(4r^2+1)^{3/2}+C