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

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

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

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

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

Solution steps

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

Simplify

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

\int \frac{1}{e ^{3 y}} d y

f (r) = 2 π r^{2}

\frac{d}{d x} ((8 x + 7) (x^{2} - 3 x))

\frac{d}{d x} (sin (2 π (1 - x)))

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