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

The integral of 1/(x^2(4+x^2)) is -1/(4x)-1/8 arctan(x/2)+C

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

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

- \frac{1}{4 x} - \frac{1}{8} arctan (\frac{x}{2}) + C

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{1}{4 ( x ^{2} + 4 )} d x = \frac{1}{8} arctan (\frac{x}{2})

= - \frac{1}{4 x} - \frac{1}{8} arctan (\frac{x}{2})

Add a constant to the solution

= - \frac{1}{4 x} - \frac{1}{8} arctan (\frac{x}{2}) + C

What is the integral of 1/(x^2(4+x^2)) ?
The integral of 1/(x^2(4+x^2)) is -1/(4x)-1/8 arctan(x/2)+C