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

The integral of x^3(4+x^2)^{3/2} is 1/35 (5x^2-8)(x^2+4)^{5/2}+C

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

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

\frac{1}{35} (5 x^{2} - 8) (x^{2} + 4)^{\frac{5}{2}} + C

Solution steps

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

Apply u-substitution

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

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

Expand

(u - 4) u^{\frac{3}{2}} : u^{\frac{5}{2}} - 4 u^{\frac{3}{2}}

= \frac{1}{2} \cdot \int u^{\frac{5}{2}} - 4 u^{\frac{3}{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{5}{2}} d u - \int 4 u^{\frac{3}{2}} d u)

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 4 u^{\frac{3}{2}} d u = \frac{8}{5} u^{\frac{5}{2}}

= \frac{1}{2} (\frac{2}{7} u^{\frac{7}{2}} - \frac{8}{5} u^{\frac{5}{2}})

Substitute back

u = 4 + x^{2}

= \frac{1}{2} (\frac{2}{7} (4 + x^{2})^{\frac{7}{2}} - \frac{8}{5} (4 + x^{2})^{\frac{5}{2}})

Simplify

\frac{1}{2} (\frac{2}{7} (4 + x^{2})^{\frac{7}{2}} - \frac{8}{5} (4 + x^{2})^{\frac{5}{2}}) : \frac{1}{35} (5 x^{2} - 8) (x^{2} + 4)^{\frac{5}{2}}

= \frac{1}{35} (5 x^{2} - 8) (x^{2} + 4)^{\frac{5}{2}}

Add a constant to the solution

= \frac{1}{35} (5 x^{2} - 8) (x^{2} + 4)^{\frac{5}{2}} + C

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What is the integral of x^3(4+x^2)^{3/2} ?
The integral of x^3(4+x^2)^{3/2} is 1/35 (5x^2-8)(x^2+4)^{5/2}+C