What is the integral of (4x^2+8x-18)/(x^4) ?

The integral of (4x^2+8x-18)/(x^4) is -4/x-4/(x^2)+6/(x^3)+C

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integral of (4x^2+8x-18)/(x^4)

\int \frac{4 x ^{2} + 8 x - 18}{x ^{4}} d x

- \frac{4}{x} - \frac{4}{x ^{2}} + \frac{6}{x ^{3}} + C

Solution steps

\int \frac{4 x ^{2} + 8 x - 18}{x ^{4}} d x

Expand

\frac{4 x ^{2} + 8 x - 18}{x ^{4}} : \frac{4}{x ^{2}} + \frac{8}{x ^{3}} - \frac{18}{x ^{4}}

= \int \frac{4}{x ^{2}} + \frac{8}{x ^{3}} - \frac{18}{x ^{4}} 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 ^{3}} d x - \int \frac{18}{x ^{4}} d x

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

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

\int \frac{18}{x ^{4}} d x = - \frac{6}{x ^{3}}

= - \frac{4}{x} - \frac{4}{x ^{2}} - (- \frac{6}{x ^{3}})

Simplify

= - \frac{4}{x} - \frac{4}{x ^{2}} + \frac{6}{x ^{3}}

Add a constant to the solution

= - \frac{4}{x} - \frac{4}{x ^{2}} + \frac{6}{x ^{3}} + C

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What is the integral of (4x^2+8x-18)/(x^4) ?
The integral of (4x^2+8x-18)/(x^4) is -4/x-4/(x^2)+6/(x^3)+C