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

The integral of sqrt(6+y)(y+2)^2 is 2/7 (y+6)^{7/2}-16/5 (y+6)^{5/2}+32/3 (y+6)^{3/2}+C

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

\int 6 + y (y + 2)^{2} d y

\frac{2}{7} (y + 6)^{\frac{7}{2}} - \frac{16}{5} (y + 6)^{\frac{5}{2}} + \frac{32}{3} (y + 6)^{\frac{3}{2}} + C

Solution steps

\int 6 + y (y + 2)^{2} d y

Expand

6 + y (y + 2)^{2} : y^{2} 6 + y + 4 y 6 + y + 4 6 + y

= \int y^{2} 6 + y + 4 y 6 + y + 4 6 + y d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int y^{2} 6 + y d y + \int 4 y 6 + y d y + \int 4 6 + y d y

\int y^{2} 6 + y d y = \frac{2}{7} (6 + y)^{\frac{7}{2}} - \frac{24}{5} (6 + y)^{\frac{5}{2}} + 24 (6 + y)^{\frac{3}{2}}

\int 4 y 6 + y d y = 4 (\frac{2}{5} (6 + y)^{\frac{5}{2}} - 4 (6 + y)^{\frac{3}{2}})

\int 4 6 + y d y = \frac{8}{3} (6 + y)^{\frac{3}{2}}

= \frac{2}{7} (6 + y)^{\frac{7}{2}} - \frac{24}{5} (6 + y)^{\frac{5}{2}} + 24 (6 + y)^{\frac{3}{2}} + 4 (\frac{2}{5} (6 + y)^{\frac{5}{2}} - 4 (6 + y)^{\frac{3}{2}}) + \frac{8}{3} (6 + y)^{\frac{3}{2}}

Simplify

\frac{2}{7} (6 + y)^{\frac{7}{2}} - \frac{24}{5} (6 + y)^{\frac{5}{2}} + 24 (6 + y)^{\frac{3}{2}} + 4 (\frac{2}{5} (6 + y)^{\frac{5}{2}} - 4 (6 + y)^{\frac{3}{2}}) + \frac{8}{3} (6 + y)^{\frac{3}{2}} : \frac{2}{7} (y + 6)^{\frac{7}{2}} - \frac{16}{5} (y + 6)^{\frac{5}{2}} + \frac{32}{3} (y + 6)^{\frac{3}{2}}

= \frac{2}{7} (y + 6)^{\frac{7}{2}} - \frac{16}{5} (y + 6)^{\frac{5}{2}} + \frac{32}{3} (y + 6)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} (y + 6)^{\frac{7}{2}} - \frac{16}{5} (y + 6)^{\frac{5}{2}} + \frac{32}{3} (y + 6)^{\frac{3}{2}} + C

\frac{d}{d x} (6 x^{4} sin (x))

\frac{\partial}{\partial x} (4 x^{2} - 8 x y^{4} + 7 a y^{5} - 3)

sin (x + 1)

\frac{d}{d x} (- \frac{1}{2} cot (x))

\frac{d y}{d x} = e^{x} y^{2}

What is the integral of sqrt(6+y)(y+2)^2 ?
The integral of sqrt(6+y)(y+2)^2 is 2/7 (y+6)^{7/2}-16/5 (y+6)^{5/2}+32/3 (y+6)^{3/2}+C