Untegrate the following \( \frac{\int(x-1)^{2}}{\left(x^{2}+1\right)^{2}} d x \)

To integrate the given expression, we can use integration by parts or substitution, but let's look at another approach. The integral is \[ \int \frac{(x - 1)^2}{(x^2 + 1)^2} \, dx. \] First, we can expand the numerator: \[ (x - 1)^2 = x^2 - 2x + 1, \] which gives: \[ \int \frac{x^2 - 2x + 1}{(x^2 + 1)^2} \, dx = \int \frac{x^2}{(x^2 + 1)^2} \, dx - 2\int \frac{x}{(x^2 + 1)^2} \, dx + \int \frac{1}{(x^2 + 1)^2} \, dx. \] Now we can tackle each term one by one: 1. For the integral \( \int \frac{1}{(x^2 + 1)^2} \, dx \), we can use the fact that it evaluates to: \[ \int \frac{1}{(x^2 + 1)^2} \, dx = \frac{1}{2} \tan^{-1}(x) + C. \] 2. For \( \int \frac{x}{(x^2 + 1)^2} \, dx \), we can use the substitution \( u = x^2 + 1 \), \( du = 2x \, dx \): \[ \int \frac{x}{(x^2 + 1)^2} \, dx = \frac{1}{2} \int \frac{1}{u^2} \, du = -\frac{1}{2u} = -\frac{1}{2(x^2 + 1)}. \] 3. Now for the term \( \int \frac{x^2}{(x^2 + 1)^2} \, dx \): We can rewrite \( x^2 = (x^2 + 1) - 1 \), then: \[ \int \frac{x^2}{(x^2 + 1)^2} \, dx = \int \frac{(x^2 + 1) - 1}{(x^2 + 1)^2} \, dx = \int \frac{1}{x^2 + 1} \, dx - \int \frac{1}{(x^2 + 1)^2} \, dx. \] The first integral gives \( \tan^{-1}(x) \) and the second provides \( \frac{1}{2} \tan^{-1}(x) \). Putting this all together: \[ \int \frac{x^2}{(x^2 + 1)^2} \, dx = \tan^{-1}(x) - \frac{1}{2} \tan^{-1}(x) = \frac{1}{2} \tan^{-1}(x). \] Finally, we combine all parts: Putting everything back into the original equation gives: \[ \int \frac{(x - 1)^2}{(x^2 + 1)^2} \, dx = \frac{1}{2} \tan^{-1}(x) - 2\left(-\frac{1}{2(x^2 + 1)}\right) + \frac{1}{2} \tan^{-1}(x). \] Simplifying this results in: \[ \frac{1}{2} \tan^{-1}(x) + \frac{1}{x^2 + 1} + \frac{1}{2} \tan^{-1}(x) + C. \] Thus, the final answer is: \[ \int \frac{(x - 1)^{2}}{(x^{2} + 1)^{2}} \, dx = \tan^{-1}(x) + \frac{1}{x^2 + 1} + C. \]