What is the integral of 1/(sqrt(t^3)) ?

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

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integral of 1/(sqrt(t^3))

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

- \frac{2}{t} + C

Solution steps

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

Apply exponent rule:

(a^{n})^{m} = a^{n \cdot m}

t^{3} = t^{\frac{3}{2}}

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

Apply the Power Rule

: - \frac{2}{t}

= - \frac{2}{t}

Add a constant to the solution

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

\frac{\partial}{\partial x} ((x) cos (x) cos (y))

y^{^{'}} = \frac{sec ( y )}{( x - 4 ) ^{2}}

(3, - 4), (5, 6)

\frac{d}{d x} (\frac{3}{x + 1})

\int cot (x) csc (x) d x

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