What is the integral of 1/(x^2)-x^2-1/3 ?

The integral of 1/(x^2)-x^2-1/3 is -1/x-(x^3)/3-1/3 x+C

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integral of 1/(x^2)-x^2-1/3

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

- \frac{1}{x} - \frac{x ^{3}}{3} - \frac{1}{3} x + C

Solution steps

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

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

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

\int \frac{1}{3} d x = \frac{1}{3} x

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

Add a constant to the solution

= - \frac{1}{x} - \frac{x ^{3}}{3} - \frac{1}{3} x + C

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What is the integral of 1/(x^2)-x^2-1/3 ?
The integral of 1/(x^2)-x^2-1/3 is -1/x-(x^3)/3-1/3 x+C