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

The integral of x^2sqrt(6x+4) is (2sqrt(2))/(2835)(135x^2-72x+32)(3x+2)^{3/2}+C

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

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

\frac{2 2}{2835} (135 x^{2} - 72 x + 32) (3 x + 2)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ( u - 4 ) ^{2}}{216} d u

= \int \frac{u ( u - 4 ) ^{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 - 4)^{2} d u

Expand

u (u - 4)^{2} : u^{\frac{5}{2}} - 8 u^{\frac{3}{2}} + 16 u

= \frac{1}{216} \cdot \int u^{\frac{5}{2}} - 8 u^{\frac{3}{2}} + 16 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 8 u^{\frac{3}{2}} d u + \int 16 u d u)

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

\int 8 u^{\frac{3}{2}} d u = \frac{16}{5} u^{\frac{5}{2}}

\int 16 u d u = \frac{32}{3} u^{\frac{3}{2}}

= \frac{1}{216} (\frac{2}{7} u^{\frac{7}{2}} - \frac{16}{5} u^{\frac{5}{2}} + \frac{32}{3} u^{\frac{3}{2}})

Substitute back

u = 6 x + 4

= \frac{1}{216} (\frac{2}{7} (6 x + 4)^{\frac{7}{2}} - \frac{16}{5} (6 x + 4)^{\frac{5}{2}} + \frac{32}{3} (6 x + 4)^{\frac{3}{2}})

Simplify

\frac{1}{216} (\frac{2}{7} (6 x + 4)^{\frac{7}{2}} - \frac{16}{5} (6 x + 4)^{\frac{5}{2}} + \frac{32}{3} (6 x + 4)^{\frac{3}{2}}) : \frac{2 2}{2835} (135 x^{2} - 72 x + 32) (3 x + 2)^{\frac{3}{2}}

= \frac{2 2}{2835} (135 x^{2} - 72 x + 32) (3 x + 2)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2 2}{2835} (135 x^{2} - 72 x + 32) (3 x + 2)^{\frac{3}{2}} + C

- \frac{3 x}{2}

\frac{d}{d x} (2 x z (x) cos (sin (x)))

\int 8 cot^{2} (x) d x

\frac{d x}{d t} = 3 (x^{2} + 1), x (\frac{π}{4}) = 1

f (x) = \frac{2 x - 3}{4 - x}

What is the integral of x^2sqrt(6x+4) ?
The integral of x^2sqrt(6x+4) is (2sqrt(2))/(2835)(135x^2-72x+32)(3x+2)^{3/2}+C