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

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

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integral of (2-1/x)^2

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

4 x - 4 ln ∣ x ∣ - \frac{1}{x} + C

Solution steps

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

Factor

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

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

Simplify

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

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

Expand

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

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

\int 4 d x = 4 x

\int \frac{4}{x} d x = 4 ln ∣ x ∣

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

= 4 x - 4 ln ∣ x ∣ - \frac{1}{x}

Add a constant to the solution

= 4 x - 4 ln ∣ x ∣ - \frac{1}{x} + C

\int_{3}^{0} (- x^{3} - 2 x^{2} + x - 1) d x

y = \frac{x}{x + 3}, (1, 0.25)

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

f (x) = x^{x^{2}}

x \to 9 lim (25 - x)

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