What is the integral of ((y-2)^2)/(sqrt(y)) ?

The integral of ((y-2)^2)/(sqrt(y)) is 2/5 y^{5/2}-8/3 y^{3/2}+8sqrt(y)+C

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integral of ((y-2)^2)/(sqrt(y))

\int \frac{( y - 2 ) ^{2}}{y} d y

\frac{2}{5} y^{\frac{5}{2}} - \frac{8}{3} y^{\frac{3}{2}} + 8 y + C

Solution steps

\int \frac{( y - 2 ) ^{2}}{y} d y

Expand

\frac{( y - 2 ) ^{2}}{y} : y^{\frac{3}{2}} - 4 y + \frac{4}{y}

= \int y^{\frac{3}{2}} - 4 y + \frac{4}{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^{\frac{3}{2}} d y - \int 4 y d y + \int \frac{4}{y} d y

\int y^{\frac{3}{2}} d y = \frac{2}{5} y^{\frac{5}{2}}

\int 4 y d y = \frac{8}{3} y^{\frac{3}{2}}

\int \frac{4}{y} d y = 8 y

= \frac{2}{5} y^{\frac{5}{2}} - \frac{8}{3} y^{\frac{3}{2}} + 8 y

Add a constant to the solution

= \frac{2}{5} y^{\frac{5}{2}} - \frac{8}{3} y^{\frac{3}{2}} + 8 y + C

- 5 t^{4}

y = arctanh (\frac{1}{x})

\frac{d}{d x} (\frac{t}{x})

y = (2 x - 5)^{\frac{1}{3}}

\int_{- 1}^{2} x d x

What is the integral of ((y-2)^2)/(sqrt(y)) ?
The integral of ((y-2)^2)/(sqrt(y)) is 2/5 y^{5/2}-8/3 y^{3/2}+8sqrt(y)+C