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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

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