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

The integral of (3x-1/x)^2 is 3x^3-6x-1/x+C

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

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

3 x^{3} - 6 x - \frac{1}{x} + C

Solution steps

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

Simplify

(3 x - \frac{1}{x})^{2} : 9 x^{2} - 6 + \frac{1}{x ^{2}}

= \int 9 x^{2} - 6 + \frac{1}{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 9 x^{2} d x - \int 6 d x + \int \frac{1}{x ^{2}} d x

\int 9 x^{2} d x = 3 x^{3}

\int 6 d x = 6 x

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

= 3 x^{3} - 6 x - \frac{1}{x}

Add a constant to the solution

= 3 x^{3} - 6 x - \frac{1}{x} + C

\frac{d}{d y} (\frac{y ^{6}}{6} + \frac{1}{16 y ^{4}})

3 + t^{5} + sin (π t)

\frac{d y}{d x} = \frac{y}{x} - cos (\frac{y}{x})

\int (x^{4} - 6 x) d x

\int \frac{sin ( 20 x )}{1 + cos ^{2} ( 20 x )} d x

What is the integral of (3x-1/x)^2 ?
The integral of (3x-1/x)^2 is 3x^3-6x-1/x+C