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

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

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

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

\frac{x ^{2}}{2} + \frac{1}{x} + C

Solution steps

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

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

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

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

Simplify

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

Add a constant to the solution

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

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