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

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

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

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

ln ∣ x ∣ + \frac{1}{x} + C

Solution steps

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

Expand

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - (- \frac{1}{x})

Simplify

= ln ∣ x ∣ + \frac{1}{x}

Add a constant to the solution

= ln ∣ x ∣ + \frac{1}{x} + C

\frac{d}{d x} (x^{2} x - 5 x)

y = 2 - e^{x}, y = 10 e^{- x} - 5

f (x) = \frac{ln ( 4 )}{3} e^{x}

y = \frac{5 x - 2}{x}

8 x^{- 2}

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