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

The integral of 1/(x^2(x^2+4)^2) is -1/(16x)-3/64 arctan(x/2)-1/128 sin(2arctan(x/2))+C

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

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

- \frac{1}{16 x} - \frac{3}{64} arctan (\frac{x}{2}) - \frac{1}{128} sin (2 arctan (\frac{x}{2})) + C

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{1}{16 ( x ^{2} + 4 )} d x = \frac{1}{32} arctan (\frac{x}{2})

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

= - \frac{1}{16 x} - \frac{1}{32} arctan (\frac{x}{2}) - \frac{1}{64} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2})))

Simplify

- \frac{1}{16 x} - \frac{1}{32} arctan (\frac{x}{2}) - \frac{1}{64} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) : - \frac{1}{16 x} - \frac{3}{64} arctan (\frac{x}{2}) - \frac{1}{128} sin (2 arctan (\frac{x}{2}))

= - \frac{1}{16 x} - \frac{3}{64} arctan (\frac{x}{2}) - \frac{1}{128} sin (2 arctan (\frac{x}{2}))

Add a constant to the solution

= - \frac{1}{16 x} - \frac{3}{64} arctan (\frac{x}{2}) - \frac{1}{128} sin (2 arctan (\frac{x}{2})) + C

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What is the integral of 1/(x^2(x^2+4)^2) ?
The integral of 1/(x^2(x^2+4)^2) is -1/(16x)-3/64 arctan(x/2)-1/128 sin(2arctan(x/2))+C