What is the integral of (8x^4-9x^3+6x^2-2x+1)/(x^2) ?

The integral of (8x^4-9x^3+6x^2-2x+1)/(x^2) is (8x^3)/3-(9x^2)/2+6x-2ln|x|-1/x+C

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integral of (8x^4-9x^3+6x^2-2x+1)/(x^2)

\int \frac{8 x ^{4} - 9 x ^{3} + 6 x ^{2} - 2 x + 1}{x ^{2}} d x

\frac{8 x ^{3}}{3} - \frac{9 x ^{2}}{2} + 6 x - 2 ln ∣ x ∣ - \frac{1}{x} + C

Solution steps

\int \frac{8 x ^{4} - 9 x ^{3} + 6 x ^{2} - 2 x + 1}{x ^{2}} d x

Expand

\frac{8 x ^{4} - 9 x ^{3} + 6 x ^{2} - 2 x + 1}{x ^{2}} : 8 x^{2} - 9 x + 6 - \frac{2}{x} + \frac{1}{x ^{2}}

= \int 8 x^{2} - 9 x + 6 - \frac{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 8 x^{2} d x - \int 9 x d x + \int 6 d x - \int \frac{2}{x} d x + \int \frac{1}{x ^{2}} d x

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

\int 9 x d x = \frac{9 x ^{2}}{2}

\int 6 d x = 6 x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

= \frac{8 x ^{3}}{3} - \frac{9 x ^{2}}{2} + 6 x - 2 ln ∣ x ∣ - \frac{1}{x}

Add a constant to the solution

= \frac{8 x ^{3}}{3} - \frac{9 x ^{2}}{2} + 6 x - 2 ln ∣ x ∣ - \frac{1}{x} + C

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What is the integral of (8x^4-9x^3+6x^2-2x+1)/(x^2) ?
The integral of (8x^4-9x^3+6x^2-2x+1)/(x^2) is (8x^3)/3-(9x^2)/2+6x-2ln|x|-1/x+C