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

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

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

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

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

Solution steps

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

Expand

\frac{2 x ^{4} - x}{x ^{3}} : 2 x - \frac{1}{x ^{2}}

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

\int 2 x d x = x^{2}

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

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

Simplify

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

Add a constant to the solution

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

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