Express $\frac{2x^3 + x^2 + x}{(x-1)(x^2 + 1)}$ in partial fractions. Differentiate $\ln(x^2 + 1)$ with respect to $x$. Hence, find $\int \frac{2x^3 + x^2 + x}{(x-1)(x^2 + 1)} \, dx$.

1. Recognise that the fraction is improper and perform long division to extract the quotient.

The degree of the numerator is 3 and the degree of the denominator is 3, so the fraction is improper.
Denominator: $(x-1)(x^2 + 1) = x^3 - x^2 + x - 1$

By long division or inspection:
$\frac{2x^3 + x^2 + x}{x^3 - x^2 + x - 1} = 2 + \frac{(2x^3 + x^2 + x) - 2(x^3 - x^2 + x - 1)}{(x-1)(x^2 + 1)}$
$= 2 + \frac{3x^2 - x + 2}{(x-1)(x^2 + 1)}$

2. Set up the partial fraction decomposition for the proper fraction part.

Let $\frac{3x^2 - x + 2}{(x-1)(x^2 + 1)} = \frac{A}{x-1} + \frac{Bx + C}{x^2 + 1}$

Multiplying by $(x-1)(x^2 + 1)$ gives:
$3x^2 - x + 2 = A(x^2 + 1) + (Bx + C)(x - 1)$

3. Solve for the constants $A$, $B$, and $C$.

Substitute $x = 1$:
$3(1)^2 - 1 + 2 = A(1^2 + 1) 4 = 2A \implies A = 2$

Substitute $x = 0$:
$2 = A(1) + C(-1) 2 = 2 - C \implies C = 0$

Compare coefficients of $x^2$:
$3 = A + B 3 = 2 + B \implies B = 1$

4. Write the final expression in partial fractions.

$\frac{2x^3 + x^2 + x}{(x-1)(x^2 + 1)} = 2 + \frac{2}{x-1} + \frac{x}{x^2 + 1}$

1. Differentiate the function using the chain rule

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

2. Simplify the derivative

$= \frac{1}{x^2 + 1} \cdot (2x) = \frac{2x}{x^2 + 1}$

1. Express the integrand using the partial fractions found in part (a)

From part (a), we have:
$\frac{2x^3 + x^2 + x}{(x-1)(x^2 + 1)} = 2 + \frac{2}{x-1} + \frac{x}{x^2+1}$

2. Set up the integral using the partial fractions

$\int \frac{2x^3 + x^2 + x}{(x-1)(x^2 + 1)} \, dx = \int \left( 2 + \frac{2}{x-1} + \frac{x}{x^2+1} \right) \, dx$

3. Integrate the first two terms

$\int 2 \, dx = 2x$
$\int \frac{2}{x-1} \, dx = 2\ln|x-1|$

4. Use the result from part (b) to integrate the last term and state the final answer

From part (b), $\frac{d}{dx} \ln(x^2+1) = \frac{2x}{x^2+1}$.
Therefore, $\int \frac{x}{x^2+1} \, dx = \frac{1}{2}\ln(x^2+1)$.

Combining all parts:
$\int \frac{2x^3 + x^2 + x}{(x-1)(x^2 + 1)} \, dx = 2x + 2\ln|x-1| + \frac{1}{2}\ln(x^2+1) + c$