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

y^{^{''}} - 8 y^{^{'}} + 17 = 0

U (x) = - 0.04 x^{2} + 240 x - 10000

\int_{0}^{3} e^{x} - x^{2} d x

\frac{\partial}{\partial x} (y ln (x^{3} + y^{4} + 3) - 9)

x \to 2 lim (x^{2} - x + 2)

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