Express $\dfrac{2x^{2}+3x}{x^{2}+3x+2}$ in partial fractions. Hence evaluate $\displaystyle \int_{1}^{3} \dfrac{2x^{2}+3x}{x^{2}+3x+2}\,dx$. Given that $y = x\ln\left(x^{2}+3x+2\right)$, find an expression for $\dfrac{dy}{dx}$. Using the results from parts (a)(ii) and (b), evaluate $\displaystyle \int_{1}^{3} \ln\left(x^{2}+3x+2\right)\,dx$.

1. Recognise the fraction is improper and perform long division to extract the polynomial part

The degree of the numerator ($2$) is equal to the degree of the denominator ($2$), so the fraction is improper.
$2x^2 + 3x = 2(x^2 + 3x + 2) - 3x - 4$
$\frac{2x^2+3x}{x^2+3x+2} = 2 + \frac{-3x-4}{x^2+3x+2}$

2. Factorise the denominator and set up the partial fraction decomposition

$x^2 + 3x + 2 = (x+1)(x+2)$
Let $\dfrac{-3x-4}{(x+1)(x+2)} = \dfrac{A}{x+1} + \dfrac{B}{x+2}$

3. Form an equation to solve for the constants

Multiplying by $(x+1)(x+2)$ gives:
$-3x - 4 = A(x+2) + B(x+1)$

4. Solve for the constants A and B using substitution

Substitute $x = -1$:
$-3(-1) - 4 = A(-1+2) \implies 3 - 4 = A \implies A = -1$
Substitute $x = -2$:
$-3(-2) - 4 = B(-2+1) \implies 6 - 4 = -B \implies B = -2$

5. Write the final expression in partial fractions

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

1. Express the integrand using the partial fractions from part (a)(i)

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

2. Integrate the expression with respect to $x$

$\displaystyle \int_{1}^{3} \left( 2 - \dfrac{1}{x+1} - \dfrac{2}{x+2} \right) dx$
$= \left[ 2x - \ln(x+1) - 2\ln(x+2) \right]_{1}^{3}$

3. Substitute the limits and simplify the expression

$= (6 - \ln 4 - 2\ln 5) - (2 - \ln 2 - 2\ln 3)$
$= 4 - \ln 4 - \ln 25 + \ln 2 + \ln 9$
$= 4 + \ln \left( \dfrac{2 \times 9}{4 \times 25} \right)$
$= 4 + \ln \dfrac{9}{50}$ (or $\approx 2.29$)

1. Apply the product rule to differentiate $y = x\ln(x^2+3x+2)$

Let $u = x$ and $v = \ln(x^2+3x+2)$.
$\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$

2. Differentiate the logarithmic term using the chain rule and combine

$\dfrac{du}{dx} = 1$
$\dfrac{dv}{dx} = \dfrac{1}{x^2+3x+2} \cdot (2x+3) = \dfrac{2x+3}{x^2+3x+2}$
$\dfrac{dy}{dx} = x\left(\dfrac{2x+3}{x^2+3x+2}\right) + \ln(x^2+3x+2)(1)$
$\dfrac{dy}{dx} = \dfrac{2x^2+3x}{x^2+3x+2} + \ln(x^2+3x+2)$

1. Use the result from part (b) to set up an integral equation

From (b), $\frac{d}{dx}[x\ln(x^2+3x+2)] = \ln(x^2+3x+2) + \frac{2x^2+3x}{x^2+3x+2}$.
Integrating both sides from $x=1$ to $x=3$:
$\int_{1}^{3} \ln(x^2+3x+2)\,dx + \int_{1}^{3} \frac{2x^2+3x}{x^2+3x+2}\,dx = \left[ x\ln(x^2+3x+2) \right]_{1}^{3}$

2. Evaluate the right-hand side and substitute the result from part (a)(ii)

$\left[ x\ln(x^2+3x+2) \right]_{1}^{3} = 3\ln(3^2+3(3)+2) - 1\ln(1^2+3(1)+2) = 3\ln 20 - \ln 6$
From (a)(ii), let $I_a = \int_{1}^{3} \frac{2x^2+3x}{x^2+3x+2}\,dx = 4 - \ln 2 - 2\ln 5 + 2\ln 3$.
$\int_{1}^{3} \ln(x^2+3x+2)\,dx = 3\ln 20 - \ln 6 - (4 - \ln 2 - 2\ln 5 + 2\ln 3)$
$= 6\ln 2 + 3\ln 5 - \ln 2 - \ln 3 - 4 + \ln 2 + 2\ln 5 - 2\ln 3$
$= 6\ln 2 + 5\ln 5 - 3\ln 3 - 4$