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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

Add a constant to the solution

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

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