What is the integral of x^3(x^2+4)^{1/2} ?

The integral of x^3(x^2+4)^{1/2} is 1/15 (3x^2-8)(x^2+4)^{3/2}+C

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integral of x^3(x^2+4)^{1/2}

\int x^{3} (x^{2} + 4)^{\frac{1}{2}} d x

\frac{1}{15} (3 x^{2} - 8) (x^{2} + 4)^{\frac{3}{2}} + C

Solution steps

\int x^{3} (x^{2} + 4)^{\frac{1}{2}} d x

Apply u-substitution

: \int \frac{( u - 4 ) u ^{\frac{1}{2}}}{2} d u

= \int \frac{( u - 4 ) u ^{\frac{1}{2}}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int (u - 4) u^{\frac{1}{2}} d u

Expand

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

= \frac{1}{2} \cdot \int u^{\frac{3}{2}} - 4 u^{\frac{1}{2}} 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}{2} (\int u^{\frac{3}{2}} d u - \int 4 u^{\frac{1}{2}} d u)

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

\int 4 u^{\frac{1}{2}} d u = \frac{8}{3} u^{\frac{3}{2}}

= \frac{1}{2} (\frac{2}{5} u^{\frac{5}{2}} - \frac{8}{3} u^{\frac{3}{2}})

Substitute back

u = x^{2} + 4

= \frac{1}{2} (\frac{2}{5} (x^{2} + 4)^{\frac{5}{2}} - \frac{8}{3} (x^{2} + 4)^{\frac{3}{2}})

Simplify

\frac{1}{2} (\frac{2}{5} (x^{2} + 4)^{\frac{5}{2}} - \frac{8}{3} (x^{2} + 4)^{\frac{3}{2}}) : \frac{1}{15} (3 x^{2} - 8) (x^{2} + 4)^{\frac{3}{2}}

= \frac{1}{15} (3 x^{2} - 8) (x^{2} + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{15} (3 x^{2} - 8) (x^{2} + 4)^{\frac{3}{2}} + C

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What is the integral of x^3(x^2+4)^{1/2} ?
The integral of x^3(x^2+4)^{1/2} is 1/15 (3x^2-8)(x^2+4)^{3/2}+C