What is the integral of (3x^2-6)/(x^4) ?

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

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integral of (3x^2-6)/(x^4)

\int \frac{3 x ^{2} - 6}{x ^{4}} d x

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

Solution steps

\int \frac{3 x ^{2} - 6}{x ^{4}} d x

Expand

\frac{3 x ^{2} - 6}{x ^{4}} : \frac{3}{x ^{2}} - \frac{6}{x ^{4}}

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

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

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

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

Simplify

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

Add a constant to the solution

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

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