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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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