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

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

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

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

\frac{x ^{4}}{4} - \frac{x ^{2}}{2} + ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{x ^{4}}{4} - \frac{x ^{2}}{2} + ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{4}}{4} - \frac{x ^{2}}{2} + ln ∣ x ∣ + C

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