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

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

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

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

\frac{4 x ^{3}}{3} - 4 x - \frac{1}{x} + C

Solution steps

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

Expand

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

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

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

\int 4 d x = 4 x

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

= \frac{4 x ^{3}}{3} - 4 x - \frac{1}{x}

Add a constant to the solution

= \frac{4 x ^{3}}{3} - 4 x - \frac{1}{x} + C

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