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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int 1 d x = x

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

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

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

Simplify

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

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

Add a constant to the solution

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

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