Express $\dfrac{8 - 3x}{(2x + 3)(x^2 + 4)}$ in partial fractions. Differentiate $\ln\left(x^2 + 4\right)$ with respect to $x$. Use your answers to parts (i) and (ii) to find $\displaystyle \int_0^1 \frac{8 - 3x}{(2x + 3)(x^2 + 4)}\,\mathrm{d}x$.

1. Set up the partial fraction decomposition form

Let $\dfrac{8 - 3x}{(2x + 3)(x^2 + 4)} = \dfrac{A}{2x + 3} + \dfrac{Bx + C}{x^2 + 4}$

2. Form the basic equation by clearing the denominator

$8 - 3x = A(x^2 + 4) + (Bx + C)(2x + 3)$

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

Let $x = -\frac{3}{2}$:
$8 - 3\left(-\frac{3}{2}\right) = A\left(\left(-\frac{3}{2}\right)^2 + 4\right)$
$8 + \frac{9}{2} = A\left(\frac{9}{4} + 4\right)$
$\frac{25}{2} = A\left(\frac{25}{4}\right)$
$A = 2$

4. Find the values of B and C by comparing coefficients

Compare coefficients of $x^2$:
$0 = A + 2B$
$0 = 2 + 2B \implies B = -1$

Compare constant terms:
$8 = 4A + 3C$
$8 = 4(2) + 3C$
$8 = 8 + 3C \implies C = 0$

5. Write the final partial fraction expression

$\dfrac{8 - 3x}{(2x + 3)(x^2 + 4)} = \dfrac{2}{2x + 3} - \dfrac{x}{x^2 + 4}$

1. Differentiate the logarithmic function using the chain rule

$\frac{\mathrm{d}}{\mathrm{d}x} \ln(x^2 + 4) = \frac{1}{x^2 + 4} \cdot \frac{\mathrm{d}}{\mathrm{d}x}(x^2 + 4)$

2. Simplify the derivative

$= \frac{1}{x^2 + 4} \cdot (2x) = \frac{2x}{x^2 + 4}$

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

From part (i), the partial fractions are:
$\frac{8 - 3x}{(2x + 3)(x^2 + 4)} = \frac{2}{2x + 3} - \frac{x}{x^2 + 4}$

From part (ii), the derivative is:
$\frac{\mathrm{d}}{\mathrm{d}x} \ln(x^2 + 4) = \frac{2x}{x^2 + 4}$

2. Rewrite the integral using the partial fractions

$\displaystyle \int_0^1 \frac{8 - 3x}{(2x + 3)(x^2 + 4)}\,\mathrm{d}x = \int_0^1 \left( \frac{2}{2x + 3} - \frac{x}{x^2 + 4} \right) \,\mathrm{d}x$

3. Integrate each term

Using the result from part (ii), $\int \frac{x}{x^2 + 4}\,\mathrm{d}x = \frac{1}{2}\ln(x^2 + 4)$.

Integrating the first term gives $\ln(2x + 3)$.

So, the integral becomes:
$\left[ \ln(2x + 3) - \frac{1}{2}\ln(x^2 + 4) \right]_0^1$

4. Substitute the limits and simplify

Substitute $x = 1$:
$\ln(5) - \frac{1}{2}\ln(5) = \frac{1}{2}\ln 5$

Substitute $x = 0$:
$\ln(3) - \frac{1}{2}\ln(4) = \ln 3 - \ln 2$

Subtracting the lower limit from the upper limit:
$\frac{1}{2}\ln 5 - (\ln 3 - \ln 2) = \frac{1}{2}\ln 5 - \ln 3 + \ln 2$