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

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

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

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

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

Solution steps

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

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

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

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

Add a constant to the solution

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

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