What is the integral of 8x^3+9x^2-1/(x^2) ?

The integral of 8x^3+9x^2-1/(x^2) is 2x^4+3x^3+1/x+C

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integral of 8x^3+9x^2-1/(x^2)

\int 8 x^{3} + 9 x^{2} - \frac{1}{x ^{2}} d x

2 x^{4} + 3 x^{3} + \frac{1}{x} + C

Solution steps

\int 8 x^{3} + 9 x^{2} - \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 8 x^{3} d x + \int 9 x^{2} d x - \int \frac{1}{x ^{2}} d x

\int 8 x^{3} d x = 2 x^{4}

\int 9 x^{2} d x = 3 x^{3}

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

= 2 x^{4} + 3 x^{3} - (- \frac{1}{x})

Simplify

= 2 x^{4} + 3 x^{3} + \frac{1}{x}

Add a constant to the solution

= 2 x^{4} + 3 x^{3} + \frac{1}{x} + C

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What is the integral of 8x^3+9x^2-1/(x^2) ?
The integral of 8x^3+9x^2-1/(x^2) is 2x^4+3x^3+1/x+C