What is the integral of x^5(8+x^3)^{3/2} ?

The integral of x^5(8+x^3)^{3/2} is 1/3 (2/7 (8+x^3)^{7/2}-16/5 (8+x^3)^{5/2})+C

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integral of x^5(8+x^3)^{3/2}

\int x^{5} (8 + x^{3})^{\frac{3}{2}} d x

\frac{1}{3} (\frac{2}{7} (8 + x^{3})^{\frac{7}{2}} - \frac{16}{5} (8 + x^{3})^{\frac{5}{2}}) + C

Solution steps

\int x^{5} (8 + x^{3})^{\frac{3}{2}} d x

Apply u-substitution

: \int \frac{( u - 8 ) u ^{\frac{3}{2}}}{3} d u

= \int \frac{( u - 8 ) u ^{\frac{3}{2}}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int (u - 8) u^{\frac{3}{2}} d u

Expand

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

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

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

\int 8 u^{\frac{3}{2}} d u = \frac{16}{5} u^{\frac{5}{2}}

= \frac{1}{3} (\frac{2}{7} u^{\frac{7}{2}} - \frac{16}{5} u^{\frac{5}{2}})

Substitute back

u = 8 + x^{3}

= \frac{1}{3} (\frac{2}{7} (8 + x^{3})^{\frac{7}{2}} - \frac{16}{5} (8 + x^{3})^{\frac{5}{2}})

Add a constant to the solution

= \frac{1}{3} (\frac{2}{7} (8 + x^{3})^{\frac{7}{2}} - \frac{16}{5} (8 + x^{3})^{\frac{5}{2}}) + C

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What is the integral of x^5(8+x^3)^{3/2} ?
The integral of x^5(8+x^3)^{3/2} is 1/3 (2/7 (8+x^3)^{7/2}-16/5 (8+x^3)^{5/2})+C