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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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