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

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

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

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

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

Solution steps

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

Apply the Power Rule

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

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

Add a constant to the solution

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

2 sin^{2} (6 x^{4})

6 x + 8

\int x \cdot e^{2} d x

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

n = 0 \sum \infty (- \frac{1}{2})^{n + 1}

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