What is the integral of (2x^5+6x^3+1)/(x^4+3x^2) ?

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

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integral of (2x^5+6x^3+1)/(x^4+3x^2)

\int \frac{2 x ^{5} + 6 x ^{3} + 1}{x ^{4} + 3 x ^{2}} d x

x^{2} - \frac{1}{3 x} - \frac{1}{3 3} arctan (\frac{x}{3}) + C

Solution steps

\int \frac{2 x ^{5} + 6 x ^{3} + 1}{x ^{4} + 3 x ^{2}} d x

Long division

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

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

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

\int \frac{1}{x ^{4} + 3 x ^{2}} d x = - \frac{1}{3 x} - \frac{1}{3 3} arctan (\frac{x}{3})

= x^{2} - \frac{1}{3 x} - \frac{1}{3 3} arctan (\frac{x}{3})

Add a constant to the solution

= x^{2} - \frac{1}{3 x} - \frac{1}{3 3} arctan (\frac{x}{3}) + C

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