What is the integral of 1/(y^{3/2)} ?

The integral of 1/(y^{3/2)} is -2/(y^{1/2)}+C

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integral of 1/(y^{3/2)}

\int \frac{1}{y ^{\frac{3}{2}}} d y

- \frac{2}{y ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{y ^{\frac{3}{2}}} d y

Apply the Power Rule

= - \frac{2}{y ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{2}{y ^{\frac{1}{2}}} + C

\frac{\partial}{\partial x} (e^{x} x y + e^{x} y^{2})

\frac{( 2 x + y + 1 ) d y}{d x} = 1

\frac{d}{d z} (\frac{a z ^{2}}{a ^{2} + z ^{2}})

\int x^{3} x^{2} - 7 d x

\int \frac{1}{x ^{3} 9 x ^{2} - 1} d x

What is the integral of 1/(y^{3/2)} ?
The integral of 1/(y^{3/2)} is -2/(y^{1/2)}+C