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

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

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

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

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

Solution steps

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

Apply the Power Rule

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

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

Add a constant to the solution

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

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 3 y = 0

\frac{5 ( s + 2 ) ^{2}}{s ( s + 1 ) ^{3}}

\int (- 25 sin (5 x - \frac{π}{2})) d x

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

f (x) = (x - 3)^{2} - 2, at x = 0

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