What is the integral from 9 to x^3 of 1/(p^2) ?

The integral from 9 to x^3 of 1/(p^2) is -1/(x^3)+1/9

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integral from 9 to x^3 of 1/(p^2)

\int_{9}^{x^{3}} \frac{1}{p ^{2}} d p

- \frac{1}{x ^{3}} + \frac{1}{9}

Solution steps

\int_{9}^{x^{3}} \frac{1}{p ^{2}} d p

Apply the Power Rule

: [- \frac{1}{p}]_{9}^{x^{3}}

= [- \frac{1}{p}]_{9}^{x^{3}}

Compute the boundaries:

- \frac{1}{x ^{3}} + \frac{1}{9}

= - \frac{1}{x ^{3}} + \frac{1}{9}

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What is the integral from 9 to x^3 of 1/(p^2) ?
The integral from 9 to x^3 of 1/(p^2) is -1/(x^3)+1/9