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)

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

\frac{1}{x} + ln ∣ x ∣ + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \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 ∣)

Simplify

= \frac{1}{x} + ln ∣ x ∣

Add a constant to the solution

= \frac{1}{x} + ln ∣ x ∣ + C

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