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

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

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

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

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

Solution steps

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

Expand

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

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

\int 3 d x = 3 x

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

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

= 3 x - (- \frac{4}{x}) - (- \frac{1}{x ^{2}})

Simplify

= 3 x + \frac{4}{x} + \frac{1}{x ^{2}}

Add a constant to the solution

= 3 x + \frac{4}{x} + \frac{1}{x ^{2}} + C

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