What is the integral of x^2sqrt(6x+6) ?

The integral of x^2sqrt(6x+6) is sqrt(6)(2/7 (x+1)^{7/2}-4/5 (x+1)^{5/2}+2/3 (x+1)^{3/2})+C

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integral of x^2sqrt(6x+6)

\int x^{2} 6 x + 6 d x

6 (\frac{2}{7} (x + 1)^{\frac{7}{2}} - \frac{4}{5} (x + 1)^{\frac{5}{2}} + \frac{2}{3} (x + 1)^{\frac{3}{2}}) + C

Solution steps

\int x^{2} 6 x + 6 d x

x^{2} 6 x + 6 = 6 x^{2} x + 1

= \int 6 x^{2} x + 1 d x

Take the constant out:

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

= 6 \cdot \int x^{2} x + 1 d x

Apply u-substitution

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

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

Expand

(u - 1)^{2} u : u^{\frac{5}{2}} - 2 u^{\frac{3}{2}} + u

= 6 \cdot \int u^{\frac{5}{2}} - 2 u^{\frac{3}{2}} + 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

= 6 (\int u^{\frac{5}{2}} d u - \int 2 u^{\frac{3}{2}} d u + \int u d u)

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

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

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

= 6 (\frac{2}{7} u^{\frac{7}{2}} - \frac{4}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = x + 1

= 6 (\frac{2}{7} (x + 1)^{\frac{7}{2}} - \frac{4}{5} (x + 1)^{\frac{5}{2}} + \frac{2}{3} (x + 1)^{\frac{3}{2}})

Add a constant to the solution

= 6 (\frac{2}{7} (x + 1)^{\frac{7}{2}} - \frac{4}{5} (x + 1)^{\frac{5}{2}} + \frac{2}{3} (x + 1)^{\frac{3}{2}}) + C

f (x) = x^{3} + x + 5, at x = - 2

(2 x^{3} - x^{2} y) + (2 x^{2} y - x^{3}) y^{^{'}} = 0

\frac{d}{d x} (arctan (\frac{x + y}{1 - x y}))

\frac{d}{d x} (4 x - 4)

\frac{\partial}{\partial y} (3 x^{2} (5 x + 7 y))

What is the integral of x^2sqrt(6x+6) ?
The integral of x^2sqrt(6x+6) is sqrt(6)(2/7 (x+1)^{7/2}-4/5 (x+1)^{5/2}+2/3 (x+1)^{3/2})+C