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

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

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

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

\frac{x ^{3}}{3} - x + \frac{2}{x} + 3 arctan (x) + C

Solution steps

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

Long division

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

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

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

\int 1 d x = x

\int \frac{x ^{2} - 2}{x ^{4} + x ^{2}} d x = \frac{2}{x} + 3 arctan (x)

= \frac{x ^{3}}{3} - x + \frac{2}{x} + 3 arctan (x)

Add a constant to the solution

= \frac{x ^{3}}{3} - x + \frac{2}{x} + 3 arctan (x) + C

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