What is the integral of (4x^3-8)/(x^5) ?

The integral of (4x^3-8)/(x^5) is -4/x+2/(x^4)+C

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integral of (4x^3-8)/(x^5)

\int \frac{4 x ^{3} - 8}{x ^{5}} d x

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

Solution steps

\int \frac{4 x ^{3} - 8}{x ^{5}} d x

Expand

\frac{4 x ^{3} - 8}{x ^{5}} : \frac{4}{x ^{2}} - \frac{8}{x ^{5}}

= \int \frac{4}{x ^{2}} - \frac{8}{x ^{5}} 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{4}{x ^{2}} d x - \int \frac{8}{x ^{5}} d x

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

\int \frac{8}{x ^{5}} d x = - \frac{2}{x ^{4}}

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

Simplify

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

Add a constant to the solution

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

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What is the integral of (4x^3-8)/(x^5) ?
The integral of (4x^3-8)/(x^5) is -4/x+2/(x^4)+C