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

The integral of (2x-3/x)^2 is (4x^3)/3-12x-9/x+C

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

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

\frac{4 x ^{3}}{3} - 12 x - \frac{9}{x} + C

Solution steps

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

Expand

(2 x - \frac{3}{x})^{2} : 4 x^{2} - 12 + \frac{9}{x ^{2}}

= \int 4 x^{2} - 12 + \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 4 x^{2} d x - \int 12 d x + \int \frac{9}{x ^{2}} d x

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

\int 12 d x = 12 x

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

= \frac{4 x ^{3}}{3} - 12 x - \frac{9}{x}

Add a constant to the solution

= \frac{4 x ^{3}}{3} - 12 x - \frac{9}{x} + C

(- 7, - 9), (- 2, - 9)

\int (x^{1.4} + 7 x^{2.5}) d x

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

f (x) = \frac{x}{x ^{2} + 4}

t^{3} y^{^{'}} + 4 t^{2} y = e^{- t}, y (- 1) = 0

What is the integral of (2x-3/x)^2 ?
The integral of (2x-3/x)^2 is (4x^3)/3-12x-9/x+C