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 ∣

Simplify

= \frac{1}{x} + ln ∣ x ∣

Add a constant to the solution

= \frac{1}{x} + ln ∣ x ∣ + C

\frac{\partial}{\partial x} (3 x^{2} y^{4} - 4)

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

\int_{- 1}^{1} \frac{1}{2 x + 3} d x

\frac{d}{d x} (\frac{1}{3} x^{(\frac{1}{2})} - x^{(\frac{3}{2})})

f (x) = \frac{1}{x}, at x = \frac{1}{4}

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