What is the integral of (x^6+3x^5-6x+1)/(x^2) ?

The integral of (x^6+3x^5-6x+1)/(x^2) is (x^5)/5+(3x^4)/4-6ln|x|-1/x+C

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integral of (x^6+3x^5-6x+1)/(x^2)

\int \frac{x ^{6} + 3 x ^{5} - 6 x + 1}{x ^{2}} d x

\frac{x ^{5}}{5} + \frac{3 x ^{4}}{4} - 6 ln ∣ x ∣ - \frac{1}{x} + C

Solution steps

\int \frac{x ^{6} + 3 x ^{5} - 6 x + 1}{x ^{2}} d x

Expand

\frac{x ^{6} + 3 x ^{5} - 6 x + 1}{x ^{2}} : x^{4} + 3 x^{3} - \frac{6}{x} + \frac{1}{x ^{2}}

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

\int x^{4} d x = \frac{x ^{5}}{5}

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

\int \frac{6}{x} d x = 6 ln ∣ x ∣

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

= \frac{x ^{5}}{5} + \frac{3 x ^{4}}{4} - 6 ln ∣ x ∣ - \frac{1}{x}

Add a constant to the solution

= \frac{x ^{5}}{5} + \frac{3 x ^{4}}{4} - 6 ln ∣ x ∣ - \frac{1}{x} + C

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What is the integral of (x^6+3x^5-6x+1)/(x^2) ?
The integral of (x^6+3x^5-6x+1)/(x^2) is (x^5)/5+(3x^4)/4-6ln|x|-1/x+C