What is the integral of 1/(9x^4+x^2) ?

The integral of 1/(9x^4+x^2) is -1/x-3arctan(3x)+C

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integral of 1/(9x^4+x^2)

\int \frac{1}{9 x ^{4} + x ^{2}} d x

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

Solution steps

\int \frac{1}{9 x ^{4} + x ^{2}} d x

Take the partial fraction of

\frac{1}{9 x ^{4} + x ^{2}} : \frac{1}{x ^{2}} - \frac{9}{9 x ^{2} + 1}

= \int \frac{1}{x ^{2}} - \frac{9}{9 x ^{2} + 1} 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{9}{9 x ^{2} + 1} d x

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

\int \frac{9}{9 x ^{2} + 1} d x = 3 arctan (3 x)

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

Add a constant to the solution

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

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What is the integral of 1/(9x^4+x^2) ?
The integral of 1/(9x^4+x^2) is -1/x-3arctan(3x)+C