What is the integral of ((2x^2-6x+1)/(x^2)) ?

The integral of ((2x^2-6x+1)/(x^2)) is 2x-6ln|x|-1/x+C

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integral of ((2x^2-6x+1)/(x^2))

\int (\frac{2 x ^{2} - 6 x + 1}{x ^{2}}) d x

2 x - 6 ln ∣ x ∣ - \frac{1}{x} + C

Solution steps

\int \frac{2 x ^{2} - 6 x + 1}{x ^{2}} d x

Expand

\frac{2 x ^{2} - 6 x + 1}{x ^{2}} : 2 - \frac{6}{x} + \frac{1}{x ^{2}}

= \int 2 - \frac{6}{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 d x - \int \frac{6}{x} d x + \int \frac{1}{x ^{2}} d x

\int 2 d x = 2 x

\int \frac{6}{x} d x = 6 ln ∣ x ∣

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

= 2 x - 6 ln ∣ x ∣ - \frac{1}{x}

Add a constant to the solution

= 2 x - 6 ln ∣ x ∣ - \frac{1}{x} + C

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What is the integral of ((2x^2-6x+1)/(x^2)) ?
The integral of ((2x^2-6x+1)/(x^2)) is 2x-6ln|x|-1/x+C