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

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

(1 + x)^{4}

\int \frac{1}{sin ( x )} d x

\frac{1 - 2 t}{1 + 3 t}

f (x) = x^{2} + 10 x + 242 x + 8

\int 3 (e^{3 x} - e^{x}) d x

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