What is the integral of 1/(x^2)-6x ?

The integral of 1/(x^2)-6x is -1/x-3x^2+C

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integral of 1/(x^2)-6x

\int \frac{1}{x ^{2}} - 6 x d x

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

Solution steps

\int \frac{1}{x ^{2}} - 6 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 6 x d x

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

\int 6 x d x = 3 x^{2}

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

Add a constant to the solution

= - \frac{1}{x} - 3 x^{2} + C