What is the integral of 1/(x^4+16x^2) ?

The integral of 1/(x^4+16x^2) is -1/(16x)-1/64 arctan(x/4)+C

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integral of 1/(x^4+16x^2)

\int \frac{1}{x ^{4} + 16 x ^{2}} d x

- \frac{1}{16 x} - \frac{1}{64} arctan (\frac{x}{4}) + C

Solution steps

\int \frac{1}{x ^{4} + 16 x ^{2}} d x

Take the partial fraction of

\frac{1}{x ^{4} + 16 x ^{2}} : \frac{1}{16 x ^{2}} - \frac{1}{16 ( x ^{2} + 16 )}

= \int \frac{1}{16 x ^{2}} - \frac{1}{16 ( x ^{2} + 16 )} 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}{16 x ^{2}} d x - \int \frac{1}{16 ( x ^{2} + 16 )} d x

\int \frac{1}{16 x ^{2}} d x = - \frac{1}{16 x}

\int \frac{1}{16 ( x ^{2} + 16 )} d x = \frac{1}{64} arctan (\frac{x}{4})

= - \frac{1}{16 x} - \frac{1}{64} arctan (\frac{x}{4})

Add a constant to the solution

= - \frac{1}{16 x} - \frac{1}{64} arctan (\frac{x}{4}) + C

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What is the integral of 1/(x^4+16x^2) ?
The integral of 1/(x^4+16x^2) is -1/(16x)-1/64 arctan(x/4)+C