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

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

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

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

- \frac{1}{x} - 2 ln ∣ x ∣ + x + C

Solution steps

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

Expand

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

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int 1 d x = x

= - \frac{1}{x} - 2 ln ∣ x ∣ + x

Add a constant to the solution

= - \frac{1}{x} - 2 ln ∣ x ∣ + x + C

f (x) = 5 + 5 x^{2} - 2 x^{3}, at x = 2

\frac{\partial}{\partial x} (\frac{G ( m , x ) m}{x ^{2}})

\int \frac{- 3 e ^{5 x}}{5 x} d x

e^{l n (c o s (x^{2}))} 2 x + 1

x \to - \infty lim (- e^{x})

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