What is the integral of (sqrt(x)+1/(sqrt(x)))(x-2/x) ?

The integral of (sqrt(x)+1/(sqrt(x)))(x-2/x) is -4sqrt(x)+4/(sqrt(x))+2/5 x^{5/2}+2/3 x^{3/2}+C

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integral of (sqrt(x)+1/(sqrt(x)))(x-2/x)

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

- 4 x + \frac{4}{x} + \frac{2}{5} x^{\frac{5}{2}} + \frac{2}{3} x^{\frac{3}{2}} + C

Solution steps

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

Expand

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

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

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

\int \frac{2}{x ^{\frac{3}{2}}} d x = - \frac{4}{x}

\int x^{\frac{3}{2}} d x = \frac{2}{5} x^{\frac{5}{2}}

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

= - 4 x^{\frac{1}{2}} - (- \frac{4}{x}) + \frac{2}{5} x^{\frac{5}{2}} + \frac{2}{3} x^{\frac{3}{2}}

Simplify

- 4 x^{\frac{1}{2}} - (- \frac{4}{x}) + \frac{2}{5} x^{\frac{5}{2}} + \frac{2}{3} x^{\frac{3}{2}} : - 4 x + \frac{4}{x} + \frac{2}{5} x^{\frac{5}{2}} + \frac{2}{3} x^{\frac{3}{2}}

= - 4 x + \frac{4}{x} + \frac{2}{5} x^{\frac{5}{2}} + \frac{2}{3} x^{\frac{3}{2}}

Add a constant to the solution

= - 4 x + \frac{4}{x} + \frac{2}{5} x^{\frac{5}{2}} + \frac{2}{3} x^{\frac{3}{2}} + C

f (x) = (5 x^{2} - 3) (x^{2} + x + 4)

\frac{\partial}{\partial y} (y x - y^{2} - x + 6 y)

\int x^{5} (3 x^{2} + 4) d x

\frac{d}{d x} (ln (tan (x^{2})))

\frac{d}{d x} (arctan (x) + x - 1)

What is the integral of (sqrt(x)+1/(sqrt(x)))(x-2/x) ?
The integral of (sqrt(x)+1/(sqrt(x)))(x-2/x) is -4sqrt(x)+4/(sqrt(x))+2/5 x^{5/2}+2/3 x^{3/2}+C