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

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

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

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

- \frac{2}{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{3}{2}}} - \frac{1}{x ^{2}}

= \int \frac{1}{x ^{\frac{3}{2}}} - \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 ^{\frac{3}{2}}} d x - \int \frac{1}{x ^{2}} d x

\int \frac{1}{x ^{\frac{3}{2}}} d x = - \frac{2}{x}

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

= - \frac{2}{x} - (- \frac{1}{x})

Simplify

= - \frac{2}{x} + \frac{1}{x}

Add a constant to the solution

= - \frac{2}{x} + \frac{1}{x} + C

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