What is the integral of ((y^4+2y^2-1)/(sqrt(y))) ?

The integral of ((y^4+2y^2-1)/(sqrt(y))) is 2/9 y^{9/2}+4/5 y^{5/2}-2sqrt(y)+C

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integral of ((y^4+2y^2-1)/(sqrt(y)))

\int (\frac{y ^{4} + 2 y ^{2} - 1}{y}) d y

\frac{2}{9} y^{\frac{9}{2}} + \frac{4}{5} y^{\frac{5}{2}} - 2 y + C

Solution steps

\int \frac{y ^{4} + 2 y ^{2} - 1}{y} d y

Expand

\frac{y ^{4} + 2 y ^{2} - 1}{y} : y^{\frac{7}{2}} + 2 y^{\frac{3}{2}} - \frac{1}{y}

= \int y^{\frac{7}{2}} + 2 y^{\frac{3}{2}} - \frac{1}{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{7}{2}} d y + \int 2 y^{\frac{3}{2}} d y - \int \frac{1}{y} d y

\int y^{\frac{7}{2}} d y = \frac{2}{9} y^{\frac{9}{2}}

\int 2 y^{\frac{3}{2}} d y = \frac{4}{5} y^{\frac{5}{2}}

\int \frac{1}{y} d y = 2 y

= \frac{2}{9} y^{\frac{9}{2}} + \frac{4}{5} y^{\frac{5}{2}} - 2 y

Add a constant to the solution

= \frac{2}{9} y^{\frac{9}{2}} + \frac{4}{5} y^{\frac{5}{2}} - 2 y + C

\frac{\partial}{\partial z} (x yz + x y^{4} + y z^{4} + z x^{4})

\frac{\partial}{\partial z} (x sin (y))

\int_{0}^{2} sin (\frac{π t}{2}) d t

\int 18 t^{8} e^{- t^{9}} d t

\frac{\partial}{\partial z} (x y sin (\frac{π z}{3}))

What is the integral of ((y^4+2y^2-1)/(sqrt(y))) ?
The integral of ((y^4+2y^2-1)/(sqrt(y))) is 2/9 y^{9/2}+4/5 y^{5/2}-2sqrt(y)+C