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

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

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

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

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

Solution steps

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

Expand

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

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

\int 6 x^{2} d x = 2 x^{3}

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

= 2 x^{3} + \frac{x ^{2}}{2} - 2 ln ∣ x ∣ - (- \frac{1}{x})

Simplify

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

Add a constant to the solution

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

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