What is the integral of x^2sqrt(3x+3) ?

The integral of x^2sqrt(3x+3) is sqrt(3)(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(3x+3)

\int x^{2} 3 x + 3 d x

3 (\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} 3 x + 3 d x

x^{2} 3 x + 3 = 3 x^{2} x + 1

= \int 3 x^{2} x + 1 d x

Take the constant out:

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

= 3 \cdot \int x^{2} x + 1 d x

Apply u-substitution

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

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

Expand

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

= 3 \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

= 3 (\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}}

= 3 (\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

= 3 (\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

= 3 (\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

\frac{d}{d x} (4 x^{2} - x)

\int 4 x (2 x^{2} + 1)^{8} d x

f (x) = x^{2} - 5, at x = 1

x^{3}, at \frac{16}{3}

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (1) = 0, y^{^{'}} (1) = 8

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