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

The integral of (x-3/x)^2 is (x^3)/3-6x-9/x+C

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integral of (x-3/x)^2

\int (x - \frac{3}{x})^{2} d x

\frac{x ^{3}}{3} - 6 x - \frac{9}{x} + C

Solution steps

\int (x - \frac{3}{x})^{2} d x

Expand

(x - \frac{3}{x})^{2} : x^{2} - 6 + \frac{9}{x ^{2}}

= \int x^{2} - 6 + \frac{9}{x ^{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^{2} d x - \int 6 d x + \int \frac{9}{x ^{2}} d x

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

\int 6 d x = 6 x

\int \frac{9}{x ^{2}} d x = - \frac{9}{x}

= \frac{x ^{3}}{3} - 6 x - \frac{9}{x}

Add a constant to the solution

= \frac{x ^{3}}{3} - 6 x - \frac{9}{x} + C

What is the integral of (x-3/x)^2 ?
The integral of (x-3/x)^2 is (x^3)/3-6x-9/x+C