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

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

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integral of (1-1/x)^2

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

x - 2 ln ∣ x ∣ - \frac{1}{x} + C

Solution steps

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

Expand

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

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

\int 1 d x = x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

= x - 2 ln ∣ x ∣ - \frac{1}{x}

Add a constant to the solution

= x - 2 ln ∣ x ∣ - \frac{1}{x} + C

n = 1 \sum \infty \frac{3 ^{n}}{4 ^{n}}

\int_{9}^{16} x d x

\frac{d}{d x} (sin (\frac{π x}{8}))

n = 0 \sum \infty \frac{1}{n ( ln ( n ) )}

f (x) = sin (x) + \frac{2}{3} cot (x)