What is the integral of x^5sqrt(x^3+6) ?

The integral of x^5sqrt(x^3+6) is 1/3 (2/5 (x^3+6)^{5/2}-4(x^3+6)^{3/2})+C

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integral of x^5sqrt(x^3+6)

\int x^{5} x^{3} + 6 d x

\frac{1}{3} (\frac{2}{5} (x^{3} + 6)^{\frac{5}{2}} - 4 (x^{3} + 6)^{\frac{3}{2}}) + C

Solution steps

\int x^{5} x^{3} + 6 d x

Apply u-substitution

: \int \frac{( u - 6 ) u}{3} d u

= \int \frac{( u - 6 ) u}{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 - 6) u d u

Expand

(u - 6) u : u^{\frac{3}{2}} - 6 u

= \frac{1}{3} \cdot \int u^{\frac{3}{2}} - 6 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}{3} (\int u^{\frac{3}{2}} d u - \int 6 u d u)

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

\int 6 u d u = 4 u^{\frac{3}{2}}

= \frac{1}{3} (\frac{2}{5} u^{\frac{5}{2}} - 4 u^{\frac{3}{2}})

Substitute back

u = x^{3} + 6

= \frac{1}{3} (\frac{2}{5} (x^{3} + 6)^{\frac{5}{2}} - 4 (x^{3} + 6)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{3} (\frac{2}{5} (x^{3} + 6)^{\frac{5}{2}} - 4 (x^{3} + 6)^{\frac{3}{2}}) + C

\frac{\partial}{\partial x} (yz - x w)

x y^{^{''}} + 5 y^{^{'}} = 0

\frac{\partial}{\partial y} (5 x y^{3})

y = x^{2} + 5 x + 2, (- 5, 2)

What is the integral of x^5sqrt(x^3+6) ?
The integral of x^5sqrt(x^3+6) is 1/3 (2/5 (x^3+6)^{5/2}-4(x^3+6)^{3/2})+C