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

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

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

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

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

Solution steps

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

Apply the Power Rule

: - \frac{2}{t ^{\frac{1}{2}}}

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

Add a constant to the solution

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

\int \frac{2}{1 + 4 x ^{2}} d x

\int sin (\frac{1}{2})

z^{3}

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

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

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