What is the integral of x^2sqrt(6x+3) ?

The integral of x^2sqrt(6x+3) is 1/(35sqrt(3))(15x^2-6x+2)(2x+1)^{3/2}+C

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integral of x^2sqrt(6x+3)

\int x^{2} 6 x + 3 d x

\frac{1}{35 3} (15 x^{2} - 6 x + 2) (2 x + 1)^{\frac{3}{2}} + C

Solution steps

\int x^{2} 6 x + 3 d x

Apply u-substitution

= \int \frac{u ( u - 3 ) ^{2}}{216} d u

Take the constant out:

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

= \frac{1}{216} \cdot \int u (u - 3)^{2} d u

Expand

u (u - 3)^{2} : u^{\frac{5}{2}} - 6 u^{\frac{3}{2}} + 9 u

= \frac{1}{216} \cdot \int u^{\frac{5}{2}} - 6 u^{\frac{3}{2}} + 9 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}{216} (\int u^{\frac{5}{2}} d u - \int 6 u^{\frac{3}{2}} d u + \int 9 u d u)

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

\int 6 u^{\frac{3}{2}} d u = \frac{12}{5} u^{\frac{5}{2}}

\int 9 u d u = 6 u^{\frac{3}{2}}

= \frac{1}{216} (\frac{2}{7} u^{\frac{7}{2}} - \frac{12}{5} u^{\frac{5}{2}} + 6 u^{\frac{3}{2}})

Substitute back

u = 6 x + 3

= \frac{1}{216} (\frac{2}{7} (6 x + 3)^{\frac{7}{2}} - \frac{12}{5} (6 x + 3)^{\frac{5}{2}} + 6 (6 x + 3)^{\frac{3}{2}})

Simplify

\frac{1}{216} (\frac{2}{7} (6 x + 3)^{\frac{7}{2}} - \frac{12}{5} (6 x + 3)^{\frac{5}{2}} + 6 (6 x + 3)^{\frac{3}{2}}) : \frac{1}{35 3} (15 x^{2} - 6 x + 2) (2 x + 1)^{\frac{3}{2}}

= \frac{1}{35 3} (15 x^{2} - 6 x + 2) (2 x + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{35 3} (15 x^{2} - 6 x + 2) (2 x + 1)^{\frac{3}{2}} + C

d er i v a t i v e f (x) = \frac{1}{5 x + 7}

t a y l or \frac{1}{( 1 + x )}

\frac{d}{d x} (3 x^{2} - 15)

n = 1 \sum \infty \frac{n}{n + 6}

x \to 0 lim ((1 - 6 x)^{\frac{2}{x}})

What is the integral of x^2sqrt(6x+3) ?
The integral of x^2sqrt(6x+3) is 1/(35sqrt(3))(15x^2-6x+2)(2x+1)^{3/2}+C