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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int u^{2} - 4 d u

= \int u^{2} - 4 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

= \int u^{2} d u - \int 4 d u

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

\int 4 d u = 4 u

= \frac{u ^{3}}{3} - 4 u

Substitute back

u = 4 + r^{2}

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

Simplify

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

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

Add a constant to the solution

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

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

x y^{^{'}} - 2 y = 12 x^{2}

\frac{d}{d x} (x^{2} + 5 x - 8)

- 18 y^{^{''}} - 14 y^{^{'}} = 0, y (3) = - 3, y^{^{'}} (3) = - 3

\int \frac{3 x}{x ^{2} - 11} d x

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