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

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

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

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

\frac{2}{3} x^{\frac{3}{2}} + \frac{2}{x ^{\frac{1}{2}}} + C

Solution steps

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

Expand

\frac{x ^{2} - 1}{x ^{\frac{3}{2}}} : x^{\frac{1}{2}} - \frac{1}{x ^{\frac{3}{2}}}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= \frac{2}{3} x^{\frac{3}{2}} - (- \frac{2}{x ^{\frac{1}{2}}})

Simplify

= \frac{2}{3} x^{\frac{3}{2}} + \frac{2}{x ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{2}{3} x^{\frac{3}{2}} + \frac{2}{x ^{\frac{1}{2}}} + C

x \to - 1 lim (\frac{4 x - 5}{3 - x})

f (x) = x^{2} - x + 19, at x = 3

\int \frac{1}{( x - 1 ) ^{\frac{1}{3}}} d x

f (x) = (x - 1) (x - 3)^{2}, at x = 4

(1 + 2 y) d x - (4 - x) d y = 0

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