Express $\frac{-x^2 + 2x + 1}{(x-1)(x^2 + 1)}$ in the form $\frac{A}{x-1} + \frac{Bx + C}{x^2 + 1}$ where $A$, $B$ and $C$ are constants. Differentiate $\ln(x^2 + 1)$ with respect to $x$. Using your results from parts (i) and (ii), find $\int \frac{-2x^2 + 4x + 2}{(x-1)(x^2 + 1)} \, dx$.

1. Set up the basic identity for the partial fractions

$\frac{-x^2 + 2x + 1}{(x-1)(x^2 + 1)} = \frac{A}{x-1} + \frac{Bx + C}{x^2 + 1}$
$-x^2 + 2x + 1 = A(x^2 + 1) + (Bx + C)(x-1)$

2. Find the value of A by substituting a suitable value for x

Let $x = 1$:
$-(1)^2 + 2(1) + 1 = A(1^2 + 1) + 0$
$2 = 2A$
$A = 1$

3. Find the value of C by substituting another value for x

Let $x = 0$:
$-(0)^2 + 2(0) + 1 = 1(0^2 + 1) + (B(0) + C)(0-1)$
$1 = 1 - C$
$C = 0$

4. Find the value of B by comparing coefficients or substituting another value for x

Compare coefficients of $x^2$:
$-1 = A + B$
$-1 = 1 + B$
$B = -2$

Therefore, $\frac{-x^2 + 2x + 1}{(x-1)(x^2 + 1)} = \frac{1}{x-1} - \frac{2x}{x^2 + 1}$

1. Differentiate the expression using the chain rule

$\frac{d}{dx}[\ln(x^2 + 1)] = \frac{1}{x^2 + 1} \cdot \frac{d}{dx}(x^2 + 1)$
$= \frac{1}{x^2 + 1} \cdot (2x)$
$= \frac{2x}{x^2 + 1}$

1. Recall or compute the results from parts (i) and (ii)

From (i), $\frac{-x^2 + 2x + 1}{(x-1)(x^2 + 1)} = \frac{1}{x-1} - \frac{2x}{x^2 + 1}$.
From (ii), $\frac{d}{dx}[\ln(x^2 + 1)] = \frac{2x}{x^2 + 1}$.

2. Express the integrand using the result from part (i)

The integrand is $\frac{-2x^2 + 4x + 2}{(x-1)(x^2 + 1)} = 2 \left( \frac{-x^2 + 2x + 1}{(x-1)(x^2 + 1)} \right)$.
So, $\int \frac{-2x^2 + 4x + 2}{(x-1)(x^2 + 1)} \, dx = 2 \int \left( \frac{1}{x-1} - \frac{2x}{x^2 + 1} \right) \, dx$.

3. Integrate each term using standard integrals and the result from part (ii)

$= 2 \left[ \ln(x-1) - \ln(x^2 + 1) \right] + c$
$= 2\ln(x-1) - 2\ln(x^2+1) + c$ (or equivalent)