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

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

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

integral

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

- \frac{1}{x} - ln ∣ x ∣ + C

Solution steps

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

Expand

\frac{1 - x}{x ^{2}} : \frac{1}{x ^{2}} - \frac{1}{x}

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

= - \frac{1}{x} - ln ∣ x ∣

Add a constant to the solution

= - \frac{1}{x} - ln ∣ x ∣ + C

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

\frac{e ^{- 1}}{x}

(x - 1)^{2}

(e^{x} + e^{- x})^{4}

\frac{1}{( 1 + a ^{2} t ^{2} )}

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