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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

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