What is the integral of x^2sqrt(4x+6) ?

The integral of x^2sqrt(4x+6) is (sqrt(2))/(35)(5x^2-6x+6)(2x+3)^{3/2}+C

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integral of x^2sqrt(4x+6)

\int x^{2} 4 x + 6 d x

\frac{2}{35} (5 x^{2} - 6 x + 6) (2 x + 3)^{\frac{3}{2}} + C

Solution steps

\int x^{2} 4 x + 6 d x

Apply u-substitution

: \int \frac{u ( u - 6 ) ^{2}}{64} d u

= \int \frac{u ( u - 6 ) ^{2}}{64} d u

Take the constant out:

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

= \frac{1}{64} \cdot \int u (u - 6)^{2} d u

Expand

u (u - 6)^{2} : u^{\frac{5}{2}} - 12 u^{\frac{3}{2}} + 36 u

= \frac{1}{64} \cdot \int u^{\frac{5}{2}} - 12 u^{\frac{3}{2}} + 36 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}{64} (\int u^{\frac{5}{2}} d u - \int 12 u^{\frac{3}{2}} d u + \int 36 u d u)

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

\int 12 u^{\frac{3}{2}} d u = \frac{24}{5} u^{\frac{5}{2}}

\int 36 u d u = 24 u^{\frac{3}{2}}

= \frac{1}{64} (\frac{2}{7} u^{\frac{7}{2}} - \frac{24}{5} u^{\frac{5}{2}} + 24 u^{\frac{3}{2}})

Substitute back

u = 4 x + 6

= \frac{1}{64} (\frac{2}{7} (4 x + 6)^{\frac{7}{2}} - \frac{24}{5} (4 x + 6)^{\frac{5}{2}} + 24 (4 x + 6)^{\frac{3}{2}})

Simplify

\frac{1}{64} (\frac{2}{7} (4 x + 6)^{\frac{7}{2}} - \frac{24}{5} (4 x + 6)^{\frac{5}{2}} + 24 (4 x + 6)^{\frac{3}{2}}) : \frac{2}{35} (5 x^{2} - 6 x + 6) (2 x + 3)^{\frac{3}{2}}

= \frac{2}{35} (5 x^{2} - 6 x + 6) (2 x + 3)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{35} (5 x^{2} - 6 x + 6) (2 x + 3)^{\frac{3}{2}} + C

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

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What is the integral of x^2sqrt(4x+6) ?
The integral of x^2sqrt(4x+6) is (sqrt(2))/(35)(5x^2-6x+6)(2x+3)^{3/2}+C