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

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

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

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

- \frac{9}{2 x ^{2}} + \frac{8 x ^{3}}{3} + C

Solution steps

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

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

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

= - \frac{9}{2 x ^{2}} + \frac{8 x ^{3}}{3}

Add a constant to the solution

= - \frac{9}{2 x ^{2}} + \frac{8 x ^{3}}{3} + C

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