What is the integral of 81x^2-18+1/(x^2) ?

The integral of 81x^2-18+1/(x^2) is 27x^3-18x-1/x+C

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integral of 81x^2-18+1/(x^2)

\int 81 x^{2} - 18 + \frac{1}{x ^{2}} d x

27 x^{3} - 18 x - \frac{1}{x} + C

Solution steps

\int 81 x^{2} - 18 + \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 81 x^{2} d x - \int 18 d x + \int \frac{1}{x ^{2}} d x

\int 81 x^{2} d x = 27 x^{3}

\int 18 d x = 18 x

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

= 27 x^{3} - 18 x - \frac{1}{x}

Add a constant to the solution

= 27 x^{3} - 18 x - \frac{1}{x} + C

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What is the integral of 81x^2-18+1/(x^2) ?
The integral of 81x^2-18+1/(x^2) is 27x^3-18x-1/x+C