What is the integral of 1/(x^{2-9)} ?

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

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integral of 1/(x^{2-9)}

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

\frac{x ^{8}}{8} + C

Solution steps

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

Apply the Power Rule

: \frac{x ^{8}}{8}

= \frac{x ^{8}}{8}

Add a constant to the solution

= \frac{x ^{8}}{8} + C

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