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

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

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

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

- \frac{x}{1 + x ^{2}} + C

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Simplify

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

Add a constant to the solution

= - \frac{x}{1 + x ^{2}} + C

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